Maharashtra Board Class 8 Science Chapter 14 Measurement and Effects of Heat Solutions

Std 8 Science Chapter 14 Measurement And Effects Of Heat Question Answer Maharashtra Board

Class 8 Science Chapter 14 Measurement And Effects Of Heat Question Answer Maharashtra Board

1. A. Whom Should I Pair With?

Question 1. Match the following.

Group 'A'	Group 'B'
1. Temperature of a healthy human body	a. 296 K
2. Boiling point of water	b. 98.6 °F
3. Room temperature	c. 0°C
4. Freezing point of water	d. 212 °F

Answer:

Group 'A'	Group 'B'
1. Temperature of a healthy human body	b. 98.6 °F
2. Boiling point of water	d. 212 °F
3. Room temperature	a. 296 K
4. Freezing point of water	c. 0°C

In simple words: This matching exercise connects common physical states and body temperatures to their standard values across different temperature scales (Fahrenheit, Celsius, and Kelvin). Understanding these conversions and typical values is fundamental to the chapter.

🎯 Exam Tip: For matching questions, it's crucial to know the precise values and corresponding units for key physical phenomena like boiling/freezing points and human body temperature. Practice conversion between scales to avoid errors.

B. Who Is Telling The Truth?

Question a. The temperature of a substance is measured in joule.
Answer: False. (The temperature of a substance is measured in °C or °F or K.)
In simple words: Temperature measures the degree of hotness or coldness, using units like Celsius, Fahrenheit, or Kelvin, while joule is a unit of energy or heat.

🎯 Exam Tip: Distinguish clearly between heat (energy) and temperature (degree of hotness/coldness). Knowing their respective units is vital for conceptual understanding.

Question b. Heat flows from an object at higher temperature to an object at lower temperature.
Answer: True.
In simple words: Heat naturally moves from hotter areas to colder areas until both reach thermal equilibrium.

🎯 Exam Tip: This is a fundamental principle of thermodynamics. Understanding the direction of heat flow is essential for problems involving heat exchange.

Question c. The joule is the unit of heat.
Answer: True.
In simple words: Heat is a form of energy, and the standard SI unit for energy is the joule.

🎯 Exam Tip: Remember that heat, being a form of energy, shares its units with other forms of energy (joule, calorie). It's important to recognize the different units used for heat.

Question d. Objects contract on heating.
Answer: False. (In general, objects expand on heating. There are some exceptions to this, you will learn about them in Standard X.)
In simple words: Generally, when objects are heated, their particles gain kinetic energy and move farther apart, causing the object to expand, not contract.

🎯 Exam Tip: Focus on the general rule of thermal expansion. While there are exceptions (like water's anomalous expansion), the default assumption for most materials is expansion upon heating.

Question e. Atoms of a solid are free.
Answer: False. (Atoms of a solid are bound to each other due to the forces acting between them.)
In simple words: Atoms in a solid are held in fixed positions by strong intermolecular forces, allowing them only to vibrate around their mean positions, unlike the free movement in gases.

🎯 Exam Tip: Understand the particle arrangement in different states of matter. Solids have fixed, tightly packed particles, while liquids and gases allow more freedom of movement.

Question f. The average kinetic energy of atoms in a hot object is less than the average kinetic energy of atoms in a cold object.
Answer: False. (The average kinetic energy of atoms in a hot object is more than the average kinetic energy of atoms in a cold object.)
In simple words: Temperature is a measure of the average kinetic energy of the particles in a substance, so a hotter object has particles with higher average kinetic energy than a colder object.

🎯 Exam Tip: This statement relates temperature directly to the kinetic energy of particles. Higher temperature means more energetic particle motion, leading to higher average kinetic energy.

C. You Will Find If You Search.

Question a. A thermometer is used to measure ……………. .
Answer: A thermometer is used to measure temperature.
In simple words: A thermometer is a tool specifically designed to quantify the degree of hotness or coldness, which is known as temperature.

🎯 Exam Tip: Remember the primary function of common laboratory instruments. A thermometer measures temperature, not heat itself.

Question b. The apparatus used to measure heat is called a ……………. .
Answer: The apparatus used to measure heat is called a calorimeter.
In simple words: A calorimeter is a device used to measure the amount of heat exchanged during a chemical reaction or physical change, typically by observing temperature changes.

🎯 Exam Tip: Differentiate between measuring temperature (thermometer) and measuring heat (calorimeter). These are distinct concepts and require different instruments.

Question c. Temperature is the measure of the ……………. kinetic energy of the atoms in a substance.
Answer: Temperature is the measure of the average kinetic energy of the atoms in a substance.
In simple words: Temperature directly reflects the average motion energy of the particles within a substance.

🎯 Exam Tip: This defines temperature at a molecular level. The term "average" is critical, as not all atoms have the same kinetic energy at any given moment.

Question d. The heat contained in a substance is the measure of the ……………. kinetic energy of the atoms in the substance.
Answer: The heat contained in a substance is the measure of the total kinetic energy of the atoms in the substance.
In simple words: Heat content represents the sum of the kinetic energies of all individual particles within a substance.

🎯 Exam Tip: Understand the difference between temperature (average kinetic energy) and heat (total kinetic energy). A larger object at the same temperature contains more heat.

Question 2. Nishigandha kept a vessel containing all the ingredients for making tea in a solar cooker. Shivani kept a similar vessel on a stove. Whose tea will be ready first and why?
Answer: Shivani's tea will be ready first.
Reason: In a given time, the amount of heat received by the vessel on a stove is far greater than that received by the vessel kept in a solar cooker.
In simple words: Shivani's tea will cook faster because a stove provides a much higher and more direct rate of heat transfer than a solar cooker, which relies on less concentrated solar energy.

🎯 Exam Tip: This question highlights the practical application of heat transfer rates. Faster cooking implies a higher rate of heat transfer, which is typical for direct flame heating compared to solar energy.

3. Write Brief Answers.

Question a. Describe a clinical thermometer. How does it differ from the thermometer used in a laboratory?
Answer: A clinical thermometer has a narrow stem and a long bulb filled with mercury (or alcohol). There is a small constriction in the stem above the bulb. When the bulb of the thermometer is held in the armpit or the mouth of a patient, the mercury (or alcohol) in the bulb rises in the stem.
When it is taken out of the patient's body, the small constriction does not allow the mercury (or alcohol) from the stem to retreat into the bulb. Thus, this arrangement enables us to read the temperature of the patient's body at ease after the removal of the thermometer from his body.
The clinical thermometer is graduated from 35 °C to 42 °C (or from 95 °F to 108 °F). At 37 °C (98.6 °F), there is a red arrow mark which indicates the temperature of a healthy person.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक क्लिनिकल थर्मामीटर को दर्शाता है जिसमें एक संकीर्ण स्टेम और पारा से भरा एक लंबा बल्ब होता है। स्टेम में बल्ब के ठीक ऊपर एक छोटा सा संकुचन होता है, जो पारा को आसानी से वापस बल्ब में जाने से रोकता है, जिससे शरीर से हटाने के बाद भी तापमान पढ़ने में आसानी होती है। इस पर 35 °C से 42 °C तक का पैमाना होता है, जिसमें 37 °C पर एक लाल निशान स्वस्थ व्यक्ति के तापमान को दर्शाता है।
The thermometer used in a laboratory has wider range and does not have constriction like a clinical thermometer.
In simple words: A clinical thermometer has a narrow range (for human body temperature) and a constriction to hold the mercury level, while a laboratory thermometer has a wider range and no constriction, allowing mercury to fall back immediately.

🎯 Exam Tip: When describing thermometers, ensure you highlight key structural differences (constriction, range) and their functional implications for specific uses (body temperature vs. general lab measurements).

Question b. What is the difference between heat and temperature?
Answer: Heat is related to the total kinetic energy of the atoms in a substance while temperature is related to the average kinetic energy of the atoms in the substance.
Heat flows from a body at higher temperature to a body at lower temperature. Thus, temperature is a quantity that determines the direction of flow of heat. It is a quantitative measure of the degree of hotness or coldness of a body.
Higher temperature does not mean higher heat content. Suppose a vessel A contains 2 litres of water at 90 °C and a vessel B contains 100 ml of water at 91 °C. Then the heat content of water in A is greater than that of water in B, but the temperature of water in B is higher than that of water in A.
Units of heat:
Heat is usually expressed in calorie, kilocalorie and joule. It can also be expressed in erg as heat is a form of energy.
In simple words: Heat is the total thermal energy transferred due to a temperature difference, while temperature is a measure of the average kinetic energy of particles, indicating the degree of hotness or coldness.

🎯 Exam Tip: Clearly define both terms and use practical examples (like the two water vessels) to illustrate their difference. Mentioning their respective units (Joules/calories for heat, °C/°F/K for temperature) is also crucial for full marks.

Question c. Explain the construction of a calorimeter. Draw the necessary figure.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक कैलोरीमीटर की संरचना को दर्शाता है, जिसमें एक आंतरिक तांबे का बर्तन होता है जो बाहरी लकड़ी के बर्तन के अंदर होता है, दोनों के बीच एक हवा का गैप होता है। इसमें एक थर्मामीटर और एक स्टिरर (घुमाने वाला) के लिए छेद वाली एक गर्मी प्रतिरोधी ढक्कन भी होती है, जिसे आंतरिक बर्तन के ऊपर एक गर्मी प्रतिरोधी रिंग से ढका जाता है ताकि गर्मी के आदान-प्रदान को कम किया जा सके।
Figure shows the construction of a calorimeter. Like a thermo flask, a calorimeter has two vessels. The inner vessel, made of copper, is (practically) thermally isolated from the surroundings. The outer vessel is made of wood and is covered with a heat resistant lid. The lid has two holes, one for the thermometer and the other for the stirrer. The inner and outer surfaces of the inner vessel are polished for minimizing exchange of heat with the surroundings by radiation. A heat resistant ring covers the inner vessel.
In simple words: A calorimeter consists of an insulated inner copper vessel placed within an outer wooden vessel, designed with a lid, thermometer, and stirrer to minimize heat exchange with the surroundings, allowing for accurate heat measurements.

🎯 Exam Tip: When explaining the construction of a calorimeter, emphasize the insulation features (double-walled, air cavity, polished surfaces) and the purpose of each component (thermometer, stirrer) in achieving thermal isolation.

Question d. Explain why rails have gaps at specific distances.
Answer: The rails expand in summer due to increase in the temperature of the atmosphere. Also, they expand due to rise in temperature caused by the friction between the rails and the wheels of the running train. If there is no gap between successive rails of a railway line, the rails would bend due to expansion. This bending and twisting of the rails would cause accidents. Hence, a gap is kept between successive rails of a railway line to make provision for their expansion in summer.
In simple words: Gaps are left between railway tracks to allow for thermal expansion during hot weather and due to friction from trains; without these gaps, the rails would buckle, leading to accidents.

🎯 Exam Tip: This question is a classic example of thermal expansion. Be sure to explain both the cause of expansion (atmospheric heat, friction) and the consequence of not leaving gaps (buckling, accidents).

Do You Know:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र रेलवे ट्रैक के दो खंडों के बीच एक स्पष्ट अंतर को दर्शाता है, जो थर्मल विस्तार के लिए जानबूझकर छोड़ी गई जगह है। यह गर्मियों में पटरियों को तापमान बढ़ने पर मुड़ने से रोकने के लिए किया जाता है, जिससे संभावित दुर्घटनाओं से बचा जा सके।
Have you seen rails? They are not continuous. A small gap is kept between them at regular intervals. This is shown in the figure. This is kept to accommodate the change in the length of the rails with change in temperature. If this gap is not kept, then the rail will get distorted due to expansion in summer which may lead to accidents.
Similar to rails, the length of bridges can also increase due to expansion in summer. The length of the 18 km long great belt bridge in Denmark increases by 4.7 m in summer. Therefore, provision in made in the construction of the bridges to accommodate this expansion.
In simple words: Gaps in railway tracks and expansion joints in bridges are crucial design elements that accommodate the thermal expansion of materials, preventing structural damage and ensuring safety.

🎯 Exam Tip: Real-world examples like railway tracks and bridges effectively illustrate the importance of understanding and accounting for thermal expansion in engineering design.

Question e. Explain with the help of formulae the expansion coefficients of liquid and gas.
Answer: 1. A liquid is held in a container. When it is heated, both the container and the liquid expand. The expansion of the container is usually very small compared to that of the liquid in it. Often, it can be ignored.
Suppose a liquid is heated so that its temperature rises by \(\Delta T\) (very small) and its volume increases from \(V_1\) to \(V_2\). Experimentally, it is found that the increase in volume, \(V_2 - V_1\), is proportional to \(V_1\) and \(\Delta T\).
Hence, \((V_2 - V_1) \propto V_1 \Delta T\).

\( \implies V_2 - V_1 = \beta V_1 \Delta T\), where \(\beta\) is a constant of proportionality called the volumetric expansion coefficient of the liquid.
\(\beta = \frac{V_2 - V_1}{V_1 \Delta T}\)
It is expressed in per °C.
We have \(V_2 = V_1 + \beta V_1 \Delta T = V_1(1 + \beta \Delta T)\).
\(\beta\) is the increase in the volume of a liquid per unit original volume per unit rise in its temperature.
2. A gas is enclosed in a container. When it is heated at constant pressure, both the container and the gas expand. Suppose a gas is heated at constant pressure so that its temperature rises by \(\Delta T\) (very small) and its volume increases from \(V_1\) to \(V_2\). Experimentally, it is found that the increase in volume, \(V_2 - V_1\) is proportional to \(V_1\) and \(\Delta T\). Hence, \((V_2 - V_1) \propto V_1 \Delta T\).

\( \implies V_2 - V_1 = \beta V_1 \Delta T\), where \(\beta\) is a constant of proportionality, the volumetric expansion coefficient, called the constant pressure expansion coefficient.
\(\beta = \frac{V_2 - V_1}{V_1 \Delta T}\)
It is expressed in per °C.
We have \(V_2 = V_1 + \beta V_1 \Delta T = V_1(1 + \beta \Delta T)\).
\(\beta\) is the increase in the volume of a gas per unit original volume per unit rise in its temperature when the pressure is kept constant.
In simple words: The volumetric expansion coefficient (\(\beta\)) for liquids and gases quantifies how much their volume changes per unit original volume for every one-degree rise in temperature, with specific formulas describing this proportionality.

🎯 Exam Tip: For problems involving expansion coefficients, always write down the given data, the relevant formula, and ensure correct units are used. Pay attention to whether it's linear, areal, or volumetric expansion and for which state of matter.

4. Solve The Following Examples.

Question a. What must be the temperature in Fahrenheit so that it will be twice its value in Celsius?
Answer: Solution:
Data: F = 2 C

\( \implies C = F/2\), F = ?
\(\frac{F - 32}{9} = \frac{C}{5}\)
\(\frac{F - 32}{9} = \frac{F/2}{5}\)

\( \implies F - 32 = \frac{9}{5} \left( \frac{F}{2} \right) = 1.8 \left( \frac{F}{2} \right) = 0.9 F\)

\( \implies F - 0.9 F = 32\)

\( \implies 0.1 F = 32\)

\( \implies F = \frac{32}{0.1} = 320 \text{ °F}\).
In simple words: To find the Fahrenheit temperature that is double its Celsius value, we use the standard conversion formula and solve for F, resulting in 320°F.

🎯 Exam Tip: For temperature conversion problems, clearly state the given condition (e.g., F = 2C) and apply the correct conversion formula. Show all algebraic steps to avoid calculation errors.

Question b. A bridge is made from 20 m long iron rods. At temperature 18 °C, the distance between two rods is 0.4 cm. Up to what temperature will the bridge be in good shape?
Answer: Solution:
Data: \(l_1 = 20 \text{ m}\), \(l_2 - l_1 = 0.4 \text{ cm} = 4 \times 10^{-3} \text{ m}\), \(T_1 = 18 \text{ °C}\), \(\lambda \text{ for iron} = 11.5 \times 10^{-6} \text{/°C}\)
\(l_2 - l_1 = \lambda l_1 \Delta T\)

\( \implies \Delta T = \frac{l_2 - l_1}{\lambda l_1} = \frac{4 \times 10^{-3} \text{ m}}{11.5 \times 10^{-6} \text{/°C} \times 20 \text{ m}}\)
\(\Delta T = \frac{400}{23} \text{ °C} \approx 17.39 \text{ °C}\)
Now, \(\Delta T = T_f - T_i \)

\( \implies T_f = T_i + \Delta T\)

\( \implies T_f = 18 \text{ °C} + 17.39 \text{ °C} = 35.39 \text{ °C}\).
The bridge will be in good shape up to 35.39 °C.
In simple words: Given the initial length, allowable expansion, and linear expansion coefficient of iron, we can calculate the maximum temperature rise the bridge can withstand before buckling, which is 35.39 °C.

🎯 Exam Tip: In thermal expansion problems, clearly list the initial conditions, the change in length, and the linear expansion coefficient. Remember to convert all units to be consistent (e.g., cm to m) before calculations.

Question c. At 15 °C the height of Eiffel Tower is 324 m. If it is made of iron, what will be the increase in length in cm, at 30 °C?
Answer: Solution:
Data: \(\Delta T = 30 \text{ °C} - 15 \text{ °C} = 15 \text{ °C}\), \(l_1 = 324 \text{ m}\), \(\lambda \text{ for iron} = 11.5 \times 10^{-6} \text{/°C}\)
\(l_2 - l_1 = \lambda l_1 \Delta T\)
\(= 11.5 \times 10^{-6} \text{/°C} \times 324 \text{ m} \times 15 \text{ °C}\)
\(= 55890 \times 10^{-6} \text{ m}\)
\(= 55890 \times 10^{-6} \times 10^2 \text{ cm}\)
\(= 55890 \times 10^{-4} \text{ cm}\)
\(= 5.589 \text{ cm (nearly } 5.6 \text{ cm)}\)
This is the increase in the length, i.e., the increase in the height of Eiffel Tower.
In simple words: By applying the formula for linear thermal expansion, we calculate that the Eiffel Tower, being made of iron, would increase in height by approximately 5.6 cm when its temperature rises from 15 °C to 30 °C.

🎯 Exam Tip: Pay close attention to unit conversions (e.g., m to cm) required in the final answer. Ensure the temperature change (\(\Delta T\)) is correctly calculated and that the expansion coefficient used is appropriate for the material.

Question d. Two substances A and B have specific heats c and 2 c respectively. If A and B are given Q and 4Q amounts of heat respectively, the change in their temperatures is the same. If the mass of A is m, what is the mass of B?
Answer: Solution:
Data: \(c(A) = c\), \(c(B) = 2c\),
\(Q(A) = Q\), \(Q(B) = 4Q\), \(\Delta T \text{ same}\),
\(m(A) = m\), \(m(B) = ?\)
\(Q = mc \Delta T \)

\( \implies \Delta T = \frac{Q}{mc}\)
As \(\Delta T\) is the same for substances A and B,
\(\frac{Q(A)}{m(A) c(A)} = \frac{Q(B)}{m(B) c(B)}\)

\( \implies \frac{Q}{mc} = \frac{4Q}{m(B) 2c}\)

\( \implies \frac{1}{m} = \frac{2}{m(B)}\)

\( \implies m(B) = 2m\).
This is the mass of B.
In simple words: To achieve the same temperature change, substance B, which has double the specific heat and receives four times the heat, must have twice the mass of substance A.

🎯 Exam Tip: For problems involving specific heat and heat exchange, write down all given quantities and use the formula \(Q = mc\Delta T\). When comparing two substances, setting their \(\Delta T\) (or Q) equal often simplifies the problem, allowing you to solve for the unknown variable.

Question e. When a substance having mass 3 kg receives 600 cal of heat, its temperature increases by 10 °C. What is the specific heat of the substance?
Answer: Solution:
Data: \(m = 3 \text{ kg} = 3000 \text{ g}\),
\(Q = 600 \text{ cal}\), \(\Delta T = 10 \text{ °C}\), \(c = ?\)
\(Q = mc \Delta T\)

\( \implies c = \frac{Q}{m \Delta T} = \frac{600 \text{ cal}}{3000 \text{ g} \times 10 \text{ °C}}\)
\(= 0.02 \text{ cal/(g.°C)}\)
This is the specific heat of the substance.
In simple words: By using the heat absorbed, mass, and temperature change, we calculate the specific heat of the substance to be 0.02 cal/(g.°C).

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., kg to g if specific heat is in cal/g.°C). Clearly state the formula and show all calculation steps, ensuring the final answer has correct units.

Can You Recall?

Question a. Which sources do we get heat from?
Answer:
1. Sun
2. earth
3. fuels like wood, coal, petrol
4. electricity
5. atomic energy
6. air.
In simple words: Heat can come from natural sources like the sun and earth's core, chemical sources like burning fuels, or man-made sources like electricity and nuclear reactions.

🎯 Exam Tip: Listing various categories (solar, geothermal, chemical, electrical, nuclear) helps demonstrate a comprehensive understanding of heat sources.

Question b. How is heat transferred?
Answer: Heat is transferred by conduction, convection and radiation.
In simple words: Heat moves through objects by direct contact (conduction), through fluids by movement of hot particles (convection), and through space by electromagnetic waves (radiation).

🎯 Exam Tip: Be ready to define and give examples for each mode of heat transfer-conduction, convection, and radiation-as they are fundamental concepts.

Question c. Which effects of heat do you know?
Answer: Expansion, change of state, rise in temperature, emission of light, burning.
In simple words: Heat can cause materials to expand, change their physical state (like melting or boiling), increase their temperature, emit light, or undergo combustion (burning).

🎯 Exam Tip: This question tests your knowledge of the common physical and chemical effects of adding heat. Listing a variety demonstrates a broader understanding.

Question d. Some effects of heat are shown in Fig. What are they?
Answer: Rise in temperature/boiling, melting, burning, expansion.
In simple words: The diagrams illustrate various effects of heat, including water boiling (rise in temperature/phase change), ice melting (phase change), a matchstick burning (chemical reaction/burning), and a hot air balloon rising (expansion of gas).

🎯 Exam Tip: When presented with diagrams, clearly identify the specific physical phenomenon depicted in each. Relate each observation to the fundamental concepts of heat and its effects.

Question e. What are potential and kinetic energies?
Answer: The energy stored in a body because of its specific state or position is called its potential energy. The energy possessed by a body because of its motion is called it's kinetic energy.
In simple words: Potential energy is stored energy due to an object's position or state, while kinetic energy is the energy an object possesses due to its motion.

🎯 Exam Tip: Ensure clear, concise definitions for both potential and kinetic energy. Providing a simple example for each (e.g., a stretched spring for potential, a moving car for kinetic) can enhance your answer.

Project:

Question a. Collect information about bimetallic strips and discuss in your class how a fire alarm is made using it.
In simple words: A bimetallic strip, made of two different metals, bends when heated due to different expansion rates; this bending can be used to complete an electrical circuit in a fire alarm, triggering an alert.

🎯 Exam Tip: For project-based questions, describe the core principle (differential expansion) and then illustrate its practical application. Focus on how the physical change in the strip is converted into an actionable output (alarm).

Class 8 Science Chapter 14 Measurement And Effects Of Heat Important Questions And Answers

Rewrite The Following Statements By Selecting The Correct Options:

Question 1. 32°F is equal to ……………. .
(a) 212 °C
(b) 212 K
(c) 273.15 K
(d) 273.15 K
Answer: (c) 273.15 K
In simple words: 32°F is the freezing point of water, which is equivalent to 0°C or 273.15 K on the Kelvin scale.

🎯 Exam Tip: Memorize key temperature equivalences like the freezing point of water (0°C, 32°F, 273.15 K) and boiling point (100°C, 212°F, 373.15 K) for quick MCQ answers.

Question 2. -40°C is equal to ……………. .
(a) -40 °F
(b) 40 °F
(c) -8°F
(d) 40K
Answer: (a) -40 °F
In simple words: -40°C is the unique temperature at which both the Celsius and Fahrenheit scales read the same value, making it equal to -40°F.

🎯 Exam Tip: This is a common trick question in temperature conversions. Knowing that -40°C = -40°F can save time in calculations.

Question 3. The boiling point of water is ……………. .
(a) 212 K
(b) 212 °F
(c) 273.15 K
(d) 32 °F
Answer: (b) 212 °F
In simple words: The boiling point of pure water under standard atmospheric pressure is 100°C, which converts to 212°F.

🎯 Exam Tip: Always specify "pure water" and "standard atmospheric pressure" when stating boiling or freezing points. Know the values across all three major temperature scales.

Question 4. Specific heat is expressed in ……………. .
(a) J/(kg-°C)
(b) kg/(J C)
(c) J/kg
(d) cal/g
Answer: (a) J/(kg-°C)
In simple words: Specific heat measures the energy required to raise the temperature of a unit mass of a substance by one degree, commonly expressed in Joules per kilogram per degree Celsius or Kelvin.

🎯 Exam Tip: Pay close attention to the units of specific heat; common alternatives include cal/(g-°C) or J/(kg-K). Ensure you use the correct unit for the given problem context.

Question 5. The freezing point of water is ……………. .
(a) 0K
(b) 212°F
(c) 32°F
(d) 0°F
Answer: (c) 32°F
In simple words: Water freezes at 0°C, which is equivalent to 32°F on the Fahrenheit scale.

🎯 Exam Tip: Master the key reference points for water on all temperature scales (freezing and boiling points) to confidently answer such multiple-choice questions.

Find The Odd One Out And Give The Reason:

Question 1. Name the biggest source of heat received by the earth.
Answer: The Sun is the biggest source of heat received by the earth.
In simple words: The Sun is the primary natural source providing all the heat to Earth.

🎯 Exam Tip: Identifying the primary heat source for Earth is a fundamental concept in physics and environmental science.

Question 2. What is the relation between the temperature in °C and the temperature in °F?
Answer: \( \frac{F-32}{9} = \frac{C}{5} \), where C denotes temperature in °C and F denotes temperature in °F.
In simple words: The formula \( (F-32)/9 = C/5 \) converts temperatures between Celsius and Fahrenheit scales.

🎯 Exam Tip: Memorize this conversion formula, as it's frequently used in temperature-related calculations and questions.

Question 3. Name the quantity expressed in cal/(g.°C).
Answer: Specific heat is expressed in cal/(g.°C).
In simple words: Cal/(g.°C) is the unit for specific heat, which tells us how much heat energy is needed to change the temperature of a substance.

🎯 Exam Tip: Knowing the units for physical quantities helps in understanding their definitions and applications. Focus on unit-quantity relationships.

Question 4. State the formula for the coefficient of linear expansion of a solid substance.
Answer: \( \lambda = (l_2 - l_1) / (l_1 \Delta T) \).
In simple words: The coefficient of linear expansion (\( \lambda \)) describes how much a material's length changes per unit original length for a one-degree temperature change.

🎯 Exam Tip: Understanding the formula for linear expansion is key to solving problems related to changes in length due to temperature variations.

Question 5. State the formula for the coefficient of areal expansion of a solid.
Answer: \( \sigma = (A_2 - A_1) / (A_1 \Delta T) \).
In simple words: The coefficient of areal expansion (\( \sigma \)) measures how much a material's surface area changes per unit original area for a one-degree temperature change.

🎯 Exam Tip: Differentiate between linear, areal, and volumetric expansion coefficients and their respective formulas for accurate problem-solving.

Question 6. State the formula for the volumetric expansion coefficient of a solid.
Answer: \( \beta = (V_2 - V_1) / (V_1 \Delta T) \).
In simple words: The volumetric expansion coefficient (\( \beta \)) indicates how much a material's volume changes per unit original volume for a one-degree temperature change.

🎯 Exam Tip: Proper recall of expansion formulas is vital. Pay attention to the subscripts and delta notations for clarity.

Question 7. State the relation between λ and σ (if AT is very small).
Answer: \( \sigma = 2 \lambda \).
In simple words: For small temperature changes, the areal expansion coefficient is approximately twice the linear expansion coefficient.

🎯 Exam Tip: This relationship is a useful approximation for many material science problems, especially when dealing with minor thermal expansions.

Question 1. How is heat generated in the Sun?
Answer: In the interior of the Sun, at the centre, hydrogen nuclei fuse together to form helium nuclei. A lot of heat is generated in this process.
In simple words: The Sun generates heat through nuclear fusion, where hydrogen atoms combine to form helium, releasing immense energy.

🎯 Exam Tip: Understanding the basic process of nuclear fusion in stars like the Sun is a key concept in astrophysics and energy generation.

Question 2. What is geothermal energy?
Answer: Heat within the interior of the earth, e.g., that coming from the molten core of the earth, is called geothermal energy.
[Note: Volcanoes, geysers, hot springs are sources of this energy.]
In simple words: Geothermal energy is the heat originating from within the Earth, often used as a renewable energy source.

🎯 Exam Tip: Being able to define different energy sources, like geothermal, is important for general science knowledge and environmental studies.

Question 3. What is atomic energy or nuclear energy?
Answer: Energy released or obtained in nuclear fission or nuclear fusion is called atomic energy or nuclear energy.
In simple words: Atomic or nuclear energy is the energy produced from the processes of splitting (fission) or combining (fusion) atomic nuclei.

🎯 Exam Tip: Distinguish between nuclear fission and fusion as processes that release atomic energy, and know their applications.

Question 4. Name three sources of chemical energy.
Answer: Fuels like wood, coal and petrol are sources of chemical energy.
In simple words: Chemical energy is stored in substances like wood, coal, and petrol, released when they undergo chemical reactions such as burning.

🎯 Exam Tip: Provide clear examples of different forms of energy, such as chemical energy, to demonstrate comprehension.

Try this:
1. Take three similar vessels. Let us call them 'A', 'B' and 'C'.
2. Fill A with hot water and B with cold water. Put some water from A and B in C.
3. Dip your right hand in A and left hand in B and keep them immersed for 2 to 3 minutes.
4. Now dip both the hands in C. What do you feel?

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र तीन समान बर्तनों (वेसल A, वेसल B, वेसल C) को दर्शाता है। वेसल A में गर्म पानी है, वेसल B में ठंडा पानी है, और वेसल C में इन दोनों के मिश्रण से बना पानी है। यह प्रयोग यह समझने के लिए है कि मनुष्य की त्वचा तापमान का सटीक माप प्रदान नहीं करती है, बल्कि सापेक्षिक संवेदना देती है।

Question 5. Even though, both the hands are dipped in water in the same vessel, i.e., water at the same temperature, your right hand will find the water to be cold while the left hand will find the water to be hot. What is the reason for this? Think about it.
Answer: The right hand finds the water cold because it loses heat to water in C. The left-hand feels the water hot because it gains heat from water in C. (This shows that we cannot determine the temperature of an object accurately by simply touching it)
In simple words: Our hands provide relative temperature sensation, not an absolute measurement; the hand previously in hot water feels the mixed water cold because it loses heat, while the hand from cold water feels it warm because it gains heat.

🎯 Exam Tip: This experiment highlights the difference between subjective perception and objective measurement of temperature. Emphasize that touch is not a reliable temperature gauge.

Question 5. State the units in which temperature is expressed.
OR
What are the units of temperature?
Answer: Temperature is expressed in °C (degree Celsius), °F (degree Fahrenheit) and K (kelvin).
In simple words: Temperature can be measured using three common scales: Celsius, Fahrenheit, and Kelvin.

🎯 Exam Tip: Listing all three standard temperature units and their full names demonstrates a comprehensive understanding of temperature measurement.

Question 6. Draw diagrams to illustrate motion of atoms in a gas and a solid. Also explain the type of motion.
Answer: Atoms of a gas are comparatively free and move at random. Figures a and b show the velocities of atoms in a gas at high and low temperature, respectively. The direction and the length of the arrows show respectively the direction and the magnitude of the velocity of the atoms. The velocity of atoms in the gas at higher temperature is larger in magnitude than the velocity of atoms in the gas at lower temperature.

ℹ️ चित्र व्याख्या (Diagram Explanation): इस चित्र में (a) गर्म गैस के अणु और (b) ठंडी गैस के अणु दिखाए गए हैं, जहाँ तीर अणुओं की गति की दिशा और वेग को दर्शाते हैं; गर्म गैस में अणु अधिक तेज़ी से चलते हैं। (c) एक ठोस में परमाणुओं की व्यवस्था को दर्शाता है, जहाँ परमाणु एक दूसरे से बंधे हुए हैं और अपनी निश्चित माध्य स्थिति के चारों ओर दोलन करते हैं, जिससे उनका वेग तापमान बढ़ने पर बढ़ता है।

The velocities of atoms in a solid are shown by arrows in Fig. (c). The atoms in a solid are tied to one another because of the forces acting between them. This is indicated by drawing springs between adjacent atoms. Because of heat, they oscillate around their fixed mean positions. The higher the temperature of a solid, the greater is their velocity of oscillation.
In simple words: Gas atoms move freely and randomly, with higher speeds at higher temperatures, while solid atoms are bound and oscillate around fixed positions, with increased oscillation speed at higher temperatures.

🎯 Exam Tip: Clearly describe the kinetic behavior of particles in different states of matter, especially how temperature affects their motion, and relate it to the given diagrams.

Try this:
1. Take two steel vessels A and B of the same size.
2. Fill some water in A and double that amount in B. Make sure that the water in both vessels is at the same temperature.
3. Raise the temperatures of water in both vessels by 10 CC using a spirit lamp. Did it take the same time to increase the temperature in the two vessels?
Answer: No.
You must have required more time to raise the temperature of water in B. This means that for the same increase in temperature, you had to give more amount of heat to B. Thus, even though the water in A and in B have the same temperature, the amount of heat in B is more than that in A.
In simple words: No, it would take more time to heat vessel B because it contains double the amount of water, meaning more heat energy is required to achieve the same temperature rise.

🎯 Exam Tip: This experiment illustrates that the amount of heat absorbed depends not only on the temperature change but also on the mass of the substance.

Question 7. How are the different units of temperature-related?
Answer: 1. The SI unit of temperature is the kelvin (K). Temperature is also expressed in °C (degree Celsius) and °F (degree Fahrenheit).
\( \frac{F-32}{9} = \frac{C}{5} \)
\( K = C + 273.15 \)

\( \implies F = \frac{9}{5}C + 32 \)

\( \implies F = \frac{9}{5}(K-273.15) + 32 \)
Here, C denotes temperature in °C, F denotes temperature in °F and K denotes temperature in K (kelvin).
[Notes: (i) The unit degree Celsius is named in honour of Anders Celsius (1701 - 44) Swedish astronomer. He devised a temperature scale in 1742.
(ii) The unit degree Fahrenheit is named in honour of Gabriel Daniel Fahrenheit (1686 - 1736) German physicist. He developed the mercury thermometer and devised the temperature scale.
(iii) The unit kelvin is named in honour of William Thomson, 1st Baron Kelvin (of Largs) (1824 - 1907) British physicist and electrical engineer. He made significant contribution in thermodynamics and electromagnetic theory. He proposed a scale of temperature now known as the Kelvin scale or thermodynamic scale.]
In simple words: Temperature units like Celsius, Fahrenheit, and Kelvin are interconnected by specific formulas, allowing conversion between them, with Kelvin being the SI unit often related to Celsius by adding 273.15.

🎯 Exam Tip: Be prepared to state the conversion formulas between Celsius, Fahrenheit, and Kelvin, and briefly explain the origin of these scales.

Question 8. What is a liquid (mercury or alcohol) thermometer?
Answer: A thermometer in which the change in the volume of a liquid (mercury or alcohol) with temperature is used for measurement of temperature is called a liquid thermometer.
In simple words: A liquid thermometer measures temperature by observing the expansion or contraction of a liquid like mercury or alcohol inside a sealed tube.

🎯 Exam Tip: Focus on the principle of thermal expansion of liquids as the basis for how these thermometers function.

Question 9. Why has mecury been replaced by alcohol in a thermometer?
Answer: Because mercury is harmful for us, it has been replaced by alcohol in a thermometer.
[Notes: (i) Mercury thermometers are still widely used in laboratories in schools and colleges.
(ii) A thermometer is a device to measure temperature. A thermometer containing mercury in its bulb is called a mercury thermometer. There are other types of thermometer such as a thermocouple thermometer, a platinum resistance thermometer, a thermistor thermometer, etc.]
In simple words: Mercury is toxic, so alcohol is often used instead in thermometers, especially for consumer use, though mercury is still found in some laboratory settings.

🎯 Exam Tip: Highlight the safety and environmental reasons for the shift from mercury to alcohol in some thermometers, while also acknowledging mercury's continued use in specific contexts.

Question 10. Describe the construction of mercury thermometer.
Answer:

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक पारा थर्मामीटर की संरचना को दर्शाता है, जिसमें एक पतली केशिका नली (capillary tube) होती है जिसके नीचे एक बल्ब में पारा भरा होता है। नली पर तापमान का पाठ्यांक (जैसे 0°C से 110°C तक) अंकित होता है, और यह तापमान मापने के लिए पारे के थर्मल विस्तार के सिद्धांत पर काम करता है।
1. For constructing a thermometer, a thick-walled capillary tube with a uniform bore is taken. The tube has a thin-walled glass bulb at one end.
2. The bulb and a small part of the tube is filled with mercury. The other end of the tube is sealed after removing air from it.
3. The tube is then calibrated and the temperature of the substance can be read with it.
[Note: The range of an alcohol thermometer is different from that of a mercury thermometer.]
In simple words: A mercury thermometer consists of a thin glass tube with a bulb at one end filled with mercury; as temperature changes, the mercury expands or contracts, moving along a calibrated scale to indicate the temperature.

🎯 Exam Tip: When describing the construction, ensure you mention the key components: thin-walled bulb, capillary tube, mercury, and the sealed, calibrated scale.

Question 11. How does a mercury thermometer work?
Answer: 1. When the bulb of the thermometer is brought in contact with the substance whose temperature is to be measured, there is an exchange of heat between the substance and the mercury in the bulb.
2. The mercury expands or contracts depending upon whether it gains heat or loses heat. Accordingly there is a rise or fall of the level of mercury in the tube of the thermometer indicating the temperature of the substance when the mercury and the substance are in thermal equilibrium.
In simple words: A mercury thermometer works by heat exchange; when placed in contact with an object, the mercury expands or contracts with temperature changes, and its level in the capillary tube shows the temperature on a calibrated scale.

🎯 Exam Tip: Explain the process clearly: heat exchange, thermal expansion/contraction of mercury, and reaching thermal equilibrium for an accurate reading.

Question 12. Why is the bore of a thermometer made very small?
Answer: The bore of a thermometer is made very small so that even a slight variation in the temperature may cause noticeable variation in the mercury level in the tube of the thermometer. As a result, the sensitivity of the thermometer increases and even small changes in the temperature can be recorded.
In simple words: The thermometer's bore is kept very narrow to make it sensitive; this ensures that even small temperature changes cause a visible rise or fall in the mercury level, allowing for precise measurements.

🎯 Exam Tip: Connect the narrow bore directly to the increased sensitivity of the thermometer, emphasizing how it allows for detection of minute temperature changes.

Question 13. Why does the bulb of a thermometer have a thin glass wall?
Answer: The bulb of a thermometer is made of a thin glass so that it can easily conduct the heat from the substance in contact with the mercury in the bulb.
In simple words: The thermometer bulb has thin glass walls to allow heat to transfer quickly and efficiently between the object being measured and the mercury inside.

🎯 Exam Tip: Focus on the role of the thin glass in facilitating rapid and efficient heat transfer, which is crucial for quick temperature readings.

Question 14. Why does a thermometer usually break at the bulb?
Answer: The bulb of a thermometer has a thin glass wall. Therefore, a thermometer usually breaks at the bulb.
In simple words: Thermometers often break at the bulb because its glass wall is intentionally made thin for better heat conduction, making it the most fragile part.

🎯 Exam Tip: Relate the thinness of the bulb's glass, necessary for function, directly to its vulnerability to breakage.

Question 15. Explain why the capillary tube of a clinical thermometer has a constriction.
Answer: When the temperature of a patient is measured with a clinical thermometer, the mercury in the bulb expands and rises in the tube. The small constriction in the thermometer prevents the mercury thread from retreating into the bulb. Thus, the patient's temperature can be read at ease after removing the thermometer from his body.
In simple words: A clinical thermometer has a constriction to trap the mercury in the tube after it's removed from the patient, allowing time to read the temperature before the mercury falls back into the bulb.

🎯 Exam Tip: Clearly state the purpose of the constriction: to hold the mercury level, enabling convenient temperature reading after removal from the body.

Question 16. Explain why a clinical thermometer should not be washed with hot water.
Answer: A clinical thermometer is constructed for the purpose of recording the temperature of the human body. Hence, its stem is such that it can allow the mercury to rise up to a temperature of about 42 °C. The temperature of hot water may be more than 42 °C. Therefore, when a clinical thermometer is washed with hot water, mercury will not have enough room for expansion and the thermometer would break.

ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र विभिन्न प्रकार के थर्मामीटरों को दर्शाता है: (a) एक सामान्य थर्मामीटर, (b) एक क्लिनिकल थर्मामीटर, (c) एक डिजिटल थर्मामीटर, और (d) एक अधिकतम-न्यूनतम थर्मामीटर। ये सभी तापमान मापने के विभिन्न तरीके और उपयोग दर्शाते हैं।

[Note: (1) A digital thermometer has a sensor that detects the heat coming out from the body directly and displays the temperature.
(2) The maximum-minimum temperature has two scales, one against each arm of the thermometer. One scale indicates the maximum temperature reached (generally during the day) and the other scale indicates the minimum temperature reached (generally during the night).]
In simple words: Clinical thermometers are designed for human body temperatures (up to ~42°C); washing them with water hotter than this limit can cause the mercury to expand beyond its capacity, leading to the thermometer breaking.

🎯 Exam Tip: Emphasize the limited temperature range of clinical thermometers and the risk of mercury expansion and breakage if exposed to temperatures beyond this range.

Question 17. Explain what happens when there is exchange of heat between two objects.
Answer: When there is an exchange of heat between a hot object and a cold object, the temperature of the hot object falls and the temperature of the cold object rises. If this system of two objects is isolated from the surroundings so that no heat enters or leaves the system, then, heat lost by the hot object = heat gained by the cold object. After some time, the average kinetic energies of the atoms in the two objects become equal, i.e., the two objects attain the same temperature.
In simple words: When two objects with different temperatures exchange heat, the hotter one cools down and the colder one warms up until both reach the same temperature, signifying thermal equilibrium.

🎯 Exam Tip: Focus on the principle of heat transfer from hot to cold, the conservation of energy (heat lost equals heat gained in an isolated system), and the concept of thermal equilibrium.

Question 18. Define specific heat.
Answer: The specific heat of an object (substance) is the amount of heat required to increase the temperature of unit mass of that substance through one degree.
In simple words: Specific heat is the amount of energy needed to raise the temperature of one unit of mass of a substance by one degree.

🎯 Exam Tip: Memorize the precise definition of specific heat, including the "unit mass" and "one degree" components, as it's a fundamental concept.

Question 19. Write the formula for specific heat. Hence, obtain its unit.
Answer: Specific heat of a substance,
\( C = \frac{Q}{m(T_f-T_i)} \), where m is the mass of the substance and Q is the amount of heat supplied to the substance to increase its temperature from Ti to Tf.

\( \implies \) Unit of specific heat = \( \frac{\text{unit of heat (energy)}}{\text{unit of mass} \times \text{unit of temperature}} \)
The SI unit of heat is the joule (J), that of mass is kg and that of temperature is kelvin (K).

\( \implies \) The SI unit of specific heat = \( \frac{J}{(kg \cdot K)} \)
[Note: Specific heat is also expressed in J/(kg°C) and cal/(g°C).]
In simple words: The formula \( C = Q/(m \Delta T) \) defines specific heat, and its SI unit is Joules per kilogram Kelvin (J/(kg·K)), showing the energy needed to heat a unit mass by one Kelvin.

🎯 Exam Tip: Provide both the formula and the derivation of the SI unit. Also, mention common alternative units like cal/(g°C) to show comprehensive knowledge.

Question 20. The specific heat of aluminium is 0.21 cal/(g°C). What do you understand by this statement?
Answer: The amount of heat required to raise the temperature of 1g of aluminium through 1 °C is 0.21 cal.
In simple words: This statement means that to increase the temperature of 1 gram of aluminum by 1 degree Celsius, 0.21 calories of heat energy are needed.

🎯 Exam Tip: When interpreting specific heat values, ensure you clearly state the mass (1g), temperature change (1°C), and the required heat energy in the given units.

Question 21. The specific heat of gold is 0.03 cal/(g°C). Express it in J/(kg°C).
Answer: 1 cal = 4.18 joules (J)
1 gram = \( 10^{-3} \) kg

\( \implies \) 1 cal/(g°C) = \( 4.18 J / (10^{-3} kg°C) \)
= 4180 J/(kg°C)

\( \implies \) 0.03 cal/(g°C) = \( 0.03 \times 4180 J/(kg°C) \)
= 125.4 J/(kg °C)
In simple words: To convert specific heat from cal/(g°C) to J/(kg°C), we multiply by the conversion factor for calories to joules and divide by the conversion factor for grams to kilograms, resulting in 125.4 J/(kg°C) for gold.

🎯 Exam Tip: Accurately perform unit conversions. Remember common conversions like calories to joules and grams to kilograms, and apply them correctly to complex units.

Question 22. Arrange the following materials in increasing order of specific heat: aluminium, gold, iron, water.
Answer: Gold, iron, aluminium, water.
In simple words: The materials, from lowest to highest specific heat, are gold, iron, aluminum, and water, indicating water requires the most energy to change its temperature.

🎯 Exam Tip: Knowledge of approximate specific heat values for common materials (especially water being high) is useful for ordering such lists.

Use your brain power!
Question 1. Why does your mother put folded cloth strips soaked in cold water on your forehead when you have high fever?
Answer: To remove heat quickly from our body and thereby lower the temperature of the body (as water has high specific heat).
In simple words: Cold wet strips are used to quickly absorb and remove excess body heat due to water's high specific heat, helping to lower fever.

🎯 Exam Tip: Relate the practical application (reducing fever) directly to the scientific principle (water's high specific heat and heat transfer).

Question 2. Why is the calorimeter made of copper?
Answer: Copper is a good conductor of heat and has low specific heat. Also copper is durable and not highly reactive.
In simple words: Copper is used for calorimeters because it quickly conducts heat to the water inside, has a low specific heat capacity (so it absorbs minimal heat itself), and is durable and stable.

🎯 Exam Tip: When asked about material choices in scientific instruments, always provide multiple relevant properties, such as thermal conductivity, specific heat, and chemical stability.

Question 23. How will you determine the specific heat of a metal using a calorimeter?
OR
Describe the experiment to determine the specific heat of iron using an iron ball, calorimeter and water.
Answer: 1. Find the mass (mᵢ) of the iron ball.
2. Find the total mass (m_c) of the calorimeter (metal container) and the stirrer.
3. Fill the calorimeter to two-thirds of its capacity with water and find its mass (m'_c) along with the stirrer. Hence, find the mass (m_w) of the water in the calorimeter (m_w = m'_c - m_c).
4. Keep the calorimeter in the wooden box and note the temperature (T₁) of the water in the calorimeter with the thermometer.
5. Suspend the iron ball in water in a beaker. Heat the beaker so that the water starts boiling. Note the temperature (T₂) of the boiling water.
6. Transfer the iron ball quickly to the calorimeter and cover the calorimeter with the lid immediately.
7. Stir the water in the calorimeter gently and continuously for uniformity of temperature and note the maximum temperature (T₃) attained by the mixture.
8. Find the specific heat capacity of iron using the following formula:
heat lost by the iron ball = heat gained by the calorimeter, stirrer and water
[assuming that there is no exchange of heat between the system (calorimeter, stirrer, water and iron ball) and the surroundings].
\( \implies m_i c_i (T_2-T_3) = (m_c c_c + m_w c_w) (T_3-T_1) \)

\( \implies c_i = \frac{(m_c c_c + m_w c_w) (T_3-T_1)}{m_i (T_2-T_3)} \)
where c_c = specific heat of the material of the calorimeter and stirrer and c_w = specific heat of water.
Hence, the specific heat of iron (cᵢ) can be determined when other quantities are known.
In simple words: To find a metal's specific heat using a calorimeter, measure the masses and initial temperatures of the metal, calorimeter, and water. Then, heat the metal, transfer it to the calorimeter, and record the final temperature of the mixture; finally, use the heat exchange formula (heat lost by metal = heat gained by calorimeter and water) to calculate the metal's specific heat.

🎯 Exam Tip: Clearly outline each step of the experimental procedure, emphasizing accurate mass and temperature measurements. Crucially, show the correct application of the principle of calorimetry (heat lost = heat gained) with the derived formula.

Question 24. Why is a calorimeter used in the study of the exchange of heat between a solid and liquid or between two liquids?
Answer: When a hot body is kept in contact with a cold one, there is an exchange of heat between the two. Hence, the temperature of the hot body decreases while that of the cold body increases till both the bodies attain the same temperature. During this process, if there is no exchange of heat between the surrounding and the bodies, the heat lost by the hot body is equal to the heat gained by the cold body.
As a calorimeter ensures that there is hardly any exchange of heat between the contents of the calorimeter and the surroundings, the calorimeter is used in the study of the exchange of heat between a solid and liquid or between two liquids.
In simple words: A calorimeter is used because it is designed to thermally isolate the contents from the surroundings, ensuring that the heat lost by one substance equals the heat gained by another within the device, allowing accurate measurement of heat exchange.

🎯 Exam Tip: The core idea here is insulation. Explain how the calorimeter minimizes heat loss/gain to/from the surroundings, making it suitable for studying heat exchange accurately.

Question 25. Explain why the inside and outside of a calorimeter are polished.
Answer: 1. A calorimeter is used for the measurement of heat. Hence, it is essential to minimize the exchange of heat between the vessel and the surroundings.
2. A polished surface is a good reflector of heat. Hence, by polishing the inside and outside of a calorimeter, the loss or gain of heat due to radiation is reduced to a considerable extent.
In simple words: Calorimeters are polished on both inner and outer surfaces because shiny surfaces are poor emitters and good reflectors of heat, which minimizes heat transfer by radiation, thereby improving the accuracy of heat measurements.

🎯 Exam Tip: Link the polished surfaces directly to the reduction of heat transfer by radiation, explaining that polished surfaces are poor absorbers and emitters, thus improving insulation.

Question 26. Explain with the help of a formula the coefficient of linear expansion of a solid.
Answer: Suppose a rod of length l₁ at temperature T₁ is heated to temperature T₂ such that \( \Delta T \) = T₂ - T₁ is very small. Let l₂ be the length of the rod at temperature T₂.
Experimentally, it is found that the increase in the length of the rod (linear expension), l₂ - l₁, is proportional to l₁ and \( \Delta T \). Therefore, \( (l_2 - l_1) \propto l_1 \Delta T \)
In simple words: The coefficient of linear expansion explains how a solid's length changes with temperature; its formula shows that the change in length is proportional to the original length and the temperature change.

🎯 Exam Tip: Start by defining linear expansion clearly, then introduce the proportionality and finally present the formula, explaining each variable.

Question 27. Define coefficient off linear explansion of a solid. Write the formula for it and obtain its unit.
Answer: Coefficient of linear expansion of a solid is defined as the increase in the length of a rod of the solid per unit initial length per unit rise in its temperature.
Coefficient of linear expansion of a solid,
\( \implies l_2 - l_1 = \lambda l_1 \Delta T \), where \( \lambda \) is the constant of proportionality, called the coefficient of linear expansion of the solid.

\( \implies \lambda = \frac{l_2-l_1}{l_1 \Delta T} \), where l₁ and l₂ are respectively the initial and final length of the rod of the solid and \( \Delta T \) is the rise in its temperature.
Unit of \( \lambda \) = \( \frac{\text{unit of length}}{\text{unit of length} \times \text{unit of temperature}} \)
It is expressed in per °C.
We have \( l_2 = l_1 + \lambda l_1 \Delta T = l_1 (1 + \lambda \Delta T) \).
In simple words: The coefficient of linear expansion quantifies how much a material's length increases per unit of its original length for every degree rise in temperature, with its unit typically being per degree Celsius (°C⁻¹) or per Kelvin (K⁻¹).

🎯 Exam Tip: Provide a precise definition, the mathematical formula with variables explained, and a clear derivation of the unit to secure full marks.

Question 28. The coefficient of linear expansion of silver is \( 18 \times 10^{-6} \) per °C. What do you understand by this statement?
Answer: If the temperature of a silver rod of length 1m is increased by 1 °C, the length of the rod increases by \( 18 \times 10^{-6} \) m.
In simple words: This statement means that for every 1°C increase in temperature, a 1-meter long silver rod will expand by \( 18 \times 10^{-6} \) meters.

🎯 Exam Tip: When explaining the meaning of a coefficient value, state the specific change in length for a unit original length and a unit temperature change.

Question 29. Explain with the help of a formula the coefficient of areal expansion of a solid.
Answer: Suppose a sheet of a solid with surface area A₁ at temperature T₁ is heated to temperature T₂ such that \( \Delta T \) = T₂ - T₁ is very small. Let A₂ be the surface area of the sheet at temperature T₂. Experimentally, it is found that the increase in the surface area of the sheet (areal expansion), A₂ - A₁, is proportional to A₁ and \( \Delta T \). Therefore, \( (A_2 - A_1) \propto A_1 \Delta T \)

\( \implies A_2 - A_1 = \sigma A_1 \Delta T \), where \( \sigma \) is the constant of proportionality, called the coefficient of areal expansion of the solid.
In simple words: The coefficient of areal expansion (\( \sigma \)) describes how a solid's surface area changes with temperature; its formula shows the area change is directly proportional to the original area and the temperature change.

🎯 Exam Tip: Define areal expansion, then show its proportionality to original area and temperature change, culminating in the formula \( \Delta A = \sigma A_1 \Delta T \).

 

Question 30. Explain with the help of a formula the volumetric expansion coefficient of a solid.

Answer:Suppose a solid with volume \(V_1\) at temperature \(T_1\) is heated to temperature \(T_2\) such that \(T_1 - T_2 - T_1\) is very small. Let \(V_2\) be the volume of the solid at temperature \(T_2\). Experimentally, it is found that the increase in the volume of the solid (volumetric expansion), \(V_2 - V_1\), is proportional to \(V_1\) and \(\Delta T\). Therefore, \((V_2 - V_1)\alpha V_1\Delta T\).
Therefore, \(V_2 - V_1 = \beta V_1\Delta T\), where \(\beta\) is the constant of proportionality, called the volumetric expansion coefficient of the solid. \[\beta = \frac{V_2-V_1}{V_1\Delta T}\] It is expressed in per °C. We have \(V_2 = V_1 + \beta V_1 \Delta T = V_1 (1 + \beta\Delta T)\). \(\beta\) is the increase in the volume of a solid per unit original volume per unit rise in its temperature. [Note: It can be shown that \(\beta = \frac{3}{2} \sigma 3\lambda\).]
In simple words: The volumetric expansion coefficient for a solid describes how much its volume changes per unit original volume for every degree Celsius rise in temperature. It is calculated by dividing the change in volume by the original volume and the temperature change.

🎯 Exam Tip: Understanding the proportionality between volume change and original volume/temperature change is crucial for solving numerical problems related to volumetric expansion.

 

Question 31. Obtain an expression for the variation of the density of a solid with temperature.

Answer:Density, \(\rho = \frac{\text{mass}}{\text{volume}}\)
Therefore, \(\rho_1 = \frac{\text{mass}}{V_1}\) and \(\rho_2 = \frac{\text{mass}}{V_2}\)
Therefore, \(\frac{\rho_1}{\rho_2} = \frac{V_2}{V_1} = 1 + \beta\Delta T\)
Therefore, \(\rho_2 = \frac{\rho_1}{1 + \beta\Delta T}\) As the temperature increases, density decreases.
In simple words: As a solid heats up, its volume generally increases, but its mass stays the same. Since density is mass divided by volume, an increase in volume with constant mass means the density decreases.

🎯 Exam Tip: Remember that density is inversely proportional to volume. This relationship is key to explaining density changes with temperature due to thermal expansion.

 

Question 1. Which use of the expansion of liquids in daily life do you know?

Answer:Use of a thermometer. The effect of heat on water is somewhat different from that for other liquids. This is called anomalous behaviour of water. We are going to learn about it in higher standard.
In simple words: The most common daily use of liquid expansion is in thermometers, where the liquid expands when heated to show temperature. Water also has a unique expansion pattern near freezing.

🎯 Exam Tip: Thermometers are a prime example of thermal expansion in liquids. Be prepared to discuss water's anomalous behavior as a special case.

 

Question 1. Using the formula density = mass/volume, explain what will be the effect of heat on the gas kept in a closed bottle.

Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र गैस पर गर्मी के प्रभाव को दर्शाता है। इसमें दो बोतलें हैं, एक में निश्चित पिस्टन (fixed piston) है और दूसरी में चलने वाला पिस्टन (movable piston) है। दोनों बोतलों के नीचे बर्नर हैं जो गैस को गर्म करते हैं। अणु गैस में अनियमित गति दिखाते हैं। The density of the gas will remain the same as there is no change on the mass and volume of the gas.
In simple words: In a closed bottle, the gas mass and volume cannot change, so heating it increases pressure but its density remains constant according to the formula.

🎯 Exam Tip: For a gas in a closed container, heating leads to an increase in pressure due to increased kinetic energy of molecules, but its overall density remains unchanged because neither mass nor volume changes.

 

Question 2. If the bottle is not closed but has a movable piston attached to its open end (see the figure), what will be the effect of heating the gas in the bottle?

Answer:The piston will move upwards as the gas expands. Therefore, the expansion of a gas is measured by keeping its pressure constant.
In simple words: When a gas with a movable piston is heated, it expands and pushes the piston upwards, demonstrating gas expansion at constant pressure.

🎯 Exam Tip: Gas expansion against a movable piston is a classic example of constant-pressure expansion, where the volume changes directly with temperature.

 

Question 1. The density of a gas decreases on heating. Which of the pictures in Fig. makes use of this?

Answer:Students should be able to answer this question.
In simple words: This question refers to an external figure, likely illustrating applications like hot air balloons where decreased gas density upon heating causes buoyancy and lift.

🎯 Exam Tip: Relate the decrease in gas density upon heating to real-world applications like hot air balloons, where heated air becomes less dense and rises, creating lift.

 

Question 1. The Celsius temperature scale:

Answer:In this case, the temperature at which pure ice melts at normal atmospheric pressure is taken as zero degree (0°C) and the temperature at which pure water boils at normal atmospheric pressure is taken as hundred degree (100 °C). The interval between them is divided into 100 equal parts. Each part corresponds to a temperature difference of 1 °C.
In simple words: The Celsius scale sets the freezing point of water at 0°C and the boiling point at 100°C, dividing the interval into 100 equal degrees.

🎯 Exam Tip: Know the fixed points (freezing and boiling of water) and the interval division for the Celsius scale. This is fundamental for temperature conversions.

 

Question 2. The Fahrenheit temperature scale:

Answer:In this case, the normal melting point of pure ice is taken as 32 °F and the normal boiling point of pure water is taken as 212 °F. The interval between these two points is divided into 180 equal parts. Each part corresponds to a temperature difference of 1°F. If C is the temperature of a body on the Celsius scale and F is the corresponding temperature on the Fahrenheit scale, the two are related by \[C = \frac{5}{9} (F-32) \text{ or } F = \frac{9}{5} C + 32.\]
In simple words: The Fahrenheit scale defines water's freezing point as 32°F and boiling point as 212°F, with 180 divisions between them, and has specific formulas for conversion with Celsius.

🎯 Exam Tip: Memorize the freezing and boiling points for Fahrenheit and the conversion formulas with Celsius, as these are frequently tested in basic thermodynamics.

 

Question 3. The Kelvin temperature scale:

Answer:In this case, the melting point of pure ice at normal atmospheric pressure is taken as 273.15 kelvin (273.15 K) and the boiling point of pure water at normal atmospheric pressure is taken as 373.15 kelvin (373.15 K). The lowest possible temperature (called the absolute zero) on this scale corresponds to -273.15 °C.
In simple words: The Kelvin scale uses absolute zero (-273.15°C) as its starting point, with water freezing at 273.15 K and boiling at 373.15 K, making it the absolute temperature scale.

🎯 Exam Tip: Understand that Kelvin is an absolute temperature scale, with 0 K representing absolute zero. Conversions between Celsius and Kelvin are simple additions/subtractions of 273.15.

 

Question 1. We heat the neck of a glass bottle in order to remove the firmly fitted stopper.

Answer:When we heat the neck of the glass bottle, it expands. Due to this, the stopper becomes loose and can be removed with ease.
In simple words: Heating the glass bottle's neck causes it to expand, making the opening wider and loosening a tight stopper, allowing for easy removal.

🎯 Exam Tip: This illustrates the principle of thermal expansion in solids. Heating causes expansion, which can be used in practical applications like loosening tight fittings.

 

Question 2. A clinical thermometer has to be jerked before re-use.

Answer:The constriction above the bulb of the clinical thermometer prevents the mercury thread once risen from falling down to the bulb. Hence, in order to force the mercury thread back into the bulb, the clinical thermometer has to be jerked before re-use.
In simple words: A clinical thermometer has a constriction that holds the mercury level after measurement, so it must be vigorously shaken or jerked to reset the mercury back into the bulb for the next use.

🎯 Exam Tip: The constriction is a unique feature of clinical thermometers designed to allow easy reading after removal from the body. Jerking helps overcome this constriction.

 

Question 3. Instead of the mercury thermometer, alcohol thermometer is used for measurement of very low temperature.

Answer:Mercury freezes at - 39 °C and turns to a solid. Hence, the mercury thermometer cannot be used to measure temperature below -39°C. The freezing point of alcohol is -117°C. Hence, the alcohol thermometer is used for measurement of very low temperature.
In simple words: Alcohol thermometers are preferred for very low temperatures because mercury freezes at -39°C, while alcohol remains liquid much further down to -117°C.

🎯 Exam Tip: The freezing point of the liquid used is the critical factor in determining the lowest temperature a thermometer can measure. Compare mercury and alcohol's freezing points.

 

Question 1. The normal armpit temperature in humans is 98.6 °F. What is this temperature in (i) degree Celsius (ii) kelvin ?

Solution:Data: F = 98.6 °F, C = ?, K = ?
(i) We use the formula: \(\frac{C}{5} = \frac{F-32}{9}\)
\(C = \frac{5}{9} (F-32)\)
\(C = \frac{5}{9} (98.6-32)\)
\(C = \frac{5}{9} (66.6)\)
\(C = \frac{333}{9}\) °C = 37 °C.
(ii) We use the formula: \(K = C + 273.15\)
\(K = 37 + 273.15 = 310.15 \text{ K}.\)
In simple words: Converting the normal human armpit temperature from Fahrenheit to Celsius and Kelvin involves using the standard temperature conversion formulas, resulting in 37°C and 310.15 K.

🎯 Exam Tip: Be proficient in temperature conversions between Fahrenheit, Celsius, and Kelvin. Pay close attention to the formula and calculation steps to avoid errors.

 

Question 2. Find the heat needed to raise the temperature of 2.5 kg of water from 30 °C to 40 °C. Write the answer in calorie as well as joule.

Solution:Data: m = 2.5 kg = 2500 g, \(\Delta T = 40 \text{ °C} - 30 \text{ °C} = 10 \text{ °C}\). Q = ? Heat needed to raise the temperature of 1 g of water through 1°C is 1 calorie.
Therefore, \(Q = 2500 \text{ g} \times 1 \text{ cal/(g°C)} \times 10 \text{ °C} = 25000 \text{ calories}\) Now, 1 calorie = 4.18 joules
Therefore, \(Q = 25000 \times 4.18 \text{ joules} = 104500 \text{ joules}\) Heat needed, Q = 25000 calories = 104500 joules.
In simple words: To calculate the heat needed, multiply the mass of water by the temperature change and the specific heat capacity (1 cal/g°C or 4.18 J/g°C), then convert between calories and joules.

🎯 Exam Tip: Remember the specific heat capacity of water (1 cal/g°C or 4.18 J/g°C) and the conversion factor between calorie and joule. Apply the formula \(Q = mc\Delta T\) accurately.

 

Question 3. If the temperature of water rises by 5 °C when 500 cal of heat is supplied to it, what is the mass of water?

Solution:Using \(Q = mc\Delta T\), where \(c\) for water is 1 cal/(g°C), we get \(m = \frac{Q}{c\Delta T}\) Mass of water \( = \frac{500 \text{ cal}}{1 \text{ cal/(g°C)} \times 5 \text{ °C}} = 100 \text{ g}.\)
In simple words: Given the heat supplied and the temperature rise, the mass of water can be found by dividing the heat by the product of specific heat capacity (1 cal/g°C) and the temperature change.

🎯 Exam Tip: This problem requires rearranging the specific heat formula \(Q = mc\Delta T\) to solve for mass, \(m = \frac{Q}{c\Delta T}\). Keep units consistent.

 

Question 4. How much heat is required to raise the temperature of 500 g of mercury from 20 °C to 100 °C? [Specific heat of mercury = 0.033 kcal/(kg °C)]

Solution:Data: m = 500 g = 0.5 kg, \(T_1 = 20 \text{ °C}\), \(T_2 = 100 \text{ °C}\), c = 0.033 kcal/(kg . °C), Q = ? Using \(Q = mc (T_2 - T_1)\) \(Q = 0.5 \text{ kg} \times 0.033 \text{ kcal/(kg . °C)} \times (100 \text{ °C} - 20 \text{ °C})\) \(Q = 0.5 \times 0.033 \times 80 \text{ kcal}\) \(Q = 0.033 \times 40 \text{ kcal}\)
Therefore, \(Q = 1.32 \text{ kcal}\) Heat required = 1.32 kcal. [Note: 1kcal/(kg-°C) = 1 cal/(g.°C)]
In simple words: To find the heat required, multiply the mass of mercury (in kg) by its specific heat capacity and the temperature difference. Ensure consistent units throughout the calculation.

🎯 Exam Tip: Pay close attention to units! Convert grams to kilograms if specific heat is given in kcal/(kg°C). Use the formula \(Q = mc\Delta T\) carefully.

 

Question 5. A certain mass of water at 84 °C is poured into an equal mass of water at 24 °C. What will be the resulting temperature of the mixture ?

Solution:Data: \(m_{hot} = m_{cold} = m\) (say), \(T_{hot} = 84 \text{ °C}\), \(T_{cold} = 24 \text{ °C}\), \(T_{mixture} = ?\) Heat lost by the hot water = heat gained by the cold water
Therefore, \(mc (T_{hot}-T_{mixture}) = mc (T_{mixture}-T_{cold})\)
\(T_{hot}-T_{mixture}=T_{mixture}-T_{cold}\)
Therefore, \(2T_{mixture} = T_{hot}+T_{cold}\) \[T_{mixture} = \frac{T_{hot}+T_{cold}}{2} = \frac{84 \text{ °C} + 24 \text{ °C}}{2}\] \[T_{mixture} = \frac{108 \text{ °C}}{2} = 54 \text{ °C}\] Resulting temperature of the mixture = 54 °C.
In simple words: When equal masses of the same substance at different temperatures are mixed, the final temperature is the average of the initial temperatures, assuming no heat loss to the surroundings.

🎯 Exam Tip: For mixing problems involving the same substance and equal masses, the final temperature is simply the average of the initial temperatures. This simplifies calculations greatly.

 

Question 6. A bucket contains 8 kg of water at 20 °C. When 4 kg of hot water is mixed with it, the temperature of the mixture becomes 40 °C. Calculate the temperature of the hot water. (Ignore the heat absorbed by the bucket.)

Solution:Data: \(m_1 = 8 \text{ kg}\), \(T_1 = 20 \text{ °C}\), \(m_2 = 4 \text{ kg}\), \(T_{mixture} = 40 \text{ °C}\), \(T_2=\) ? Heat lost by the hot water = heat gained by the cold water (ignoring the heat absorbed by the bucket)
Therefore, \(m_2c (T_2 – T_{mixture}) = m_1c (T_{mixture} – T_1)\)
Since 'c' (specific heat of water) is common, it cancels out: \(4 \text{ kg} \times (T_2 – 40\text{°C}) = 8 \text{ kg} \times (40\text{°C} – 20\text{°C})\)
\(4(T_2 - 40\text{°C}) = 8 \times 20\text{°C}\)
\(4T_2 - 160\text{°C} = 160\text{°C}\)
\(4T_2 = 320\text{°C}\)
Therefore, \(T_2 = \frac{320\text{°C}}{4} = 80 \text{ °C}\) Temperature of the hot water = 80 °C.
In simple words: By applying the principle of conservation of heat (heat lost equals heat gained), we can set up an equation relating the masses, specific heats, and temperature changes of the mixed water to find the unknown initial temperature of the hot water.

🎯 Exam Tip: In mixing problems, always ensure that the specific heat capacity 'c' cancels out if the same substance is involved. Carefully set up the heat lost = heat gained equation.

 

Question 7. A blacksmith plunges a 2 kg horseshoe at 400 °C into 1 kg of water at 20 °C. Find the maximum temperature of the water. [Specific heat of iron = 0.11 kcal/(kg-°C)]

Solution:Data: \(m_{iron} = 2 \text{ kg}\), \(c_{iron} = 0.11 \text{ kcal/(kg.°C)}\), \(T_{iron\_initial} = 400 \text{ °C}\), \(m_{water} = 1 \text{ kg}\), \(c_{water} = 1 \text{ kcal/(kg.°C)}\), \(T_{water\_initial} = 20 \text{ °C}\), \(T_{final} = ?\) Heat lost by the horseshoe = heat gained by the water
Therefore, \(m_{iron} c_{iron} (T_{iron\_initial} - T_{final}) = m_{water} c_{water} (T_{final} – T_{water\_initial})\)
\(2 \text{ kg} \times 0.11 \text{ kcal/(kg.°C)} \times (400 \text{ °C} – T_{final})\) \( = 1 \text{ kg} \times 1 \text{ kcal/(kg.°C)} \times (T_{final} – 20 \text{ °C})\)
\(0.22 \times (400 – T_{final}) = T_{final} – 20\)
\(88 - 0.22T_{final} = T_{final} – 20\)
\(88 + 20 = T_{final} + 0.22T_{final}\)
\(108 = 1.22T_{final}\)
Therefore, \(T_{final} = \frac{108}{1.22}\) °C \(\approx\) 88.52 °C Maximum temperature of the water = 88.52 °C.
In simple words: The final temperature of the water is found by equating the heat lost by the hot horseshoe to the heat gained by the water, considering their respective masses, specific heats, and temperature changes.

🎯 Exam Tip: For problems involving different substances, ensure you use the correct specific heat capacity for each material. Set up the "heat lost = heat gained" equation carefully and solve for the final equilibrium temperature.

 

Question 8. A copper sphere of mass 500 g is heated to 100 °C and then introduced into a copper calorimeter containing 100 g of water at 20 °C. Find the maximum temperature of the mixture, if the mass of the calorimeter is 100 g and the specific heat of the calorimeter is 0.1 cal/(g.°C).

Solution:Data: \(m_{sphere} = 500 \text{ g}\), \(c_{sphere} = 0.1 \text{ cal/(g.°C)}\), \(T_{sphere\_initial} = 100 \text{ °C}\), \(m_{water} = 100 \text{ g}\), \(c_{water} = 1 \text{ cal/(g.°C)}\), \(T_{water\_initial} = 20\text{°C}\), \(m_{calorimeter} = 100 \text{ g}\), \(c_{calorimeter} = 0.1 \text{ cal/(g.°C)}\), \(T_{calorimeter\_initial} = 20 \text{ °C}\), \(T_{final} = ?\) Heat lost by the sphere = heat gained by the water + heat gained by the calorimeter.
Therefore, \(m_{sphere} c_{sphere} (T_{sphere\_initial} – T_{final}) = m_{water} c_{water} (T_{final} – T_{water\_initial}) + m_{calorimeter} c_{calorimeter} (T_{final} – T_{calorimeter\_initial})\)
\(500 \text{ g} \times 0.1 \text{ cal/(g.°C)} \times (100 \text{ °C} – T_{final})\) \( = 100 \text{ g} \times 1 \text{ cal/(g.°C)} \times (T_{final} – 20 \text{ °C}) + 100 \text{ g} \times 0.1 \text{ cal/(g.°C)} \times (T_{final} – 20 \text{ °C})\)
\(50 (100 – T_{final}) = 100 (T_{final} – 20) + 10 (T_{final} – 20)\)
\(50 (100 – T_{final}) = 110 (T_{final} – 20)\)
\(5000 - 50T_{final} = 110T_{final} - 2200\)
\(5000 + 2200 = 110T_{final} + 50T_{final}\)
\(7200 = 160T_{final}\)
Therefore, \(T_{final} = \frac{7200}{160}\) °C = 45 °C Maximum temperature of the mixture = 45 °C.
In simple words: The final temperature of the mixture is calculated by applying the principle that total heat lost by the hot sphere equals the total heat gained by both the water and the calorimeter, considering their individual masses, specific heats, and initial temperatures.

🎯 Exam Tip: For calorimetry problems, ensure you account for all components that gain or lose heat. Be meticulous with distributing the specific heat and temperature changes for each material in the heat balance equation.

 

Question 9. A metal rod 1.8 m long, increases in length by 1.4 mm, when heated from 0 °C to 50 °C. Find the coefficient of linear expansion of the metal.

Solution:Data: \(l_1 = 1.8 \text{ m}\), \(\Delta L = l_2 – l_1 = 1.4 \text{ mm} = 1.4 \times 10^{-3} \text{ m}.\) \(T_1 = 0 \text{ °C}\), \(T_2 = 50 \text{ °C}.\) The formula for linear expansion is \(\Delta L = \lambda l_1 (T_2-T_1)\)
Therefore, The coefficient of linear expansion of the metal is \[\lambda = \frac{\Delta L}{l_1 (T_2-T_1)}\] \[\lambda = \frac{1.4 \times 10^{-3} \text{m}}{(1.8 \text{ m}) (50 \text{ °C} -0\text{ °C})}\] \[\lambda = \frac{1.4 \times 10^{-3}}{1.8 \times 50} = \frac{1.4 \times 10^{-3}}{90} \approx 1.556 \times 10^{-5} \text{ °C}^{-1}.\]
In simple words: The coefficient of linear expansion is found by dividing the change in length by the product of the original length and the temperature change.

🎯 Exam Tip: Ensure all length measurements are in consistent units (e.g., meters) and temperature in °C. The formula for linear expansion coefficient \(\lambda = \frac{\Delta L}{L_0 \Delta T}\) is fundamental here.

 

Question 10. A thin metal disc of surface area 500 cm\(^2\) at 0 °C is heated to 40 °C. Find the increase in the surface area of the disc. (\(\sigma = 4 \times 10^{-5}\) °C\(^{-1}\))

Solution:Data: \(A_1 = 500 \text{ cm}^2\), \(T_1 = 0 \text{ °C}\), \(T_2 = 40 \text{ °C}\), \(\sigma = 4 \times 10^{-5} \text{ °C}^{-1}\), \(\Delta A = A_2 - A_1 = ?\) The increase in the surface area of the disc is \(\Delta A = A_1\sigma(T_2 – T_1)\) \(\Delta A = (500 \text{ cm}^2) (4 \times 10^{-5} \text{ °C}^{-1}) (40 \text{ °C} – 0 \text{ °C})\) \(\Delta A = 500 \times 4 \times 10^{-5} \times 40 = 0.8 \text{ cm}^2.\)
In simple words: The increase in the surface area of the disc is calculated by multiplying its original area, the coefficient of areal expansion, and the change in temperature.

🎯 Exam Tip: The formula for areal expansion is \(\Delta A = A_0 \sigma \Delta T\). Remember to use consistent units for area and temperature. The coefficient of areal expansion (\(\sigma\)) is usually given or can be related to linear expansion (\(\sigma \approx 2\lambda\)).

 

Question 11. The surface area of a metal plate is 2.4 × 10\(^{-2}\) m\(^2\) at 20 °C. When the plate is heated to 185 °C, its area increases by 0.8 cm\(^2\). Find the coefficient of areal expansion of the metal.

Solution:Data: \(A_1 = 2.4 \times 10^{-2} \text{ m}^2\), \(T_1 = 20 \text{ °C}\), \(T_2 = 185 \text{ °C}\), \(\Delta A = 0.8 \text{ cm}^2 = 0.8 \times 10^{-4} \text{ m}^2\), \(\sigma = ?\) We use the formula \(\Delta A = A_1 \sigma(T_2 – T_1)\)
Therefore, The coefficient of areal expansion of the metal is \[\sigma = \frac{\Delta A}{A_1 (T_2-T_1)}\] \[\sigma = \frac{0.8 \times 10^{-4} \text{ m}^2}{(2.4 \times 10^{-2} \text{ m}^2) (185 \text{ °C} - 20 \text{ °C})}\] \[\sigma = \frac{0.8 \times 10^{-4}}{(2.4 \times 10^{-2}) \times 165} \text{ °C}^{-1}\] \[\sigma = \frac{0.8 \times 10^{-2}}{2.4 \times 165} \text{ °C}^{-1} = \frac{1}{3 \times 165} \times 10^{-2} \text{ °C}^{-1} = \frac{1}{495} \times 10^{-2} \text{ °C}^{-1} \approx 2.020 \times 10^{-5} \text{ °C}^{-1}.\]
In simple words: To determine the coefficient of areal expansion, divide the measured increase in surface area by the product of the original surface area and the total change in temperature.

🎯 Exam Tip: Always convert all units to be consistent (e.g., cm² to m²). Carefully apply the formula \(\sigma = \frac{\Delta A}{A_0 \Delta T}\) and perform calculations with scientific notation accurately.

 

Question 12. A lead bullet has a volume of 25 cm\(^3\) at 0 °C, and 25.21 cm\(^3\) at 90 °C. Find the volumetric expansion coefficient of lead.

Solution:Data: \(V_1 = 25 \text{ cm}^3\), \(T_1 = 0 \text{ °C}\), \(V_2 = 25.21 \text{ cm}^3\), \(T_2 = 90 \text{ °C}\) \(\Delta T = T_2 - T_1 = 90 \text{ °C} – 0 \text{ °C} = 90 \text{ °C}\), \(\beta = ?\) We use the formula: \[\beta = \frac{V_2-V_1}{V_1 (T_2-T_1)}\] \[\beta = \frac{25.21 \text{ cm}^3 - 25 \text{ cm}^3}{(25 \text{ cm}^3) (90 \text{ °C})}\] \[\beta = \frac{0.21}{25 \times 90} \text{ °C}^{-1} = \frac{0.21}{2250} \text{ °C}^{-1} \approx 9.333 \times 10^{-5} \text{ °C}^{-1}.\] The volumetric expansion coefficient of lead is \(9.333 \times 10^{-5}\) °C\(^{-1}\).
In simple words: The volumetric expansion coefficient for lead is found by dividing its change in volume by the product of its original volume and the temperature change it undergoes.

🎯 Exam Tip: The formula for volumetric expansion coefficient is \(\beta = \frac{\Delta V}{V_0 \Delta T}\). Pay attention to consistent units for volume and temperature and ensure accurate calculation with decimals.

 

Question 1. The temperature of a body is 30 °C. Express it in (i) degree Fahrenheit (ii) kelvin.

Answer:86 °F, 303.15 K
In simple words: To convert 30°C to Fahrenheit, use the formula \(F = \frac{9}{5}C + 32\), and to convert to Kelvin, add 273.15 to the Celsius temperature.

🎯 Exam Tip: Practice these basic temperature conversions frequently. Remember that K = °C + 273.15 and F = (9/5)°C + 32 are essential formulas for these types of questions.

 

Question 2. The temperature of a body is 283.15 K. Express it in °C and °F.

Answer:10 °C, 50 °F
In simple words: To convert Kelvin to Celsius, subtract 273.15. Then, use the Celsius to Fahrenheit conversion formula.

🎯 Exam Tip: Always start by converting Kelvin to Celsius (K - 273.15 = °C) as it's a direct relationship, and then proceed to Fahrenheit if required, using the Celsius-Fahrenheit formula.

 

Question 3. The temperature of a body is 68 °F. Express it in °C and K.

Answer:20 °C, 293.15 K
In simple words: Convert Fahrenheit to Celsius first using \(C = \frac{5}{9}(F-32)\), then convert the Celsius value to Kelvin by adding 273.15.

🎯 Exam Tip: When converting from Fahrenheit, always calculate Celsius first, as it acts as a bridge to the Kelvin scale. Double-check your arithmetic, especially with subtraction in the Fahrenheit to Celsius conversion.

 

Question 4. Find the heat needed to raise the temperature of 5 kg of water from 20 °C to 25 °C. Write the answer in calorie as well as joule.

Answer:25 × 10\(^3\) cal, 1.045 × 10\(^5\) J
In simple words: Calculate the heat required using the specific heat of water (1 cal/g°C or 4.18 J/g°C), converting mass to grams for calories or using kilograms for joules, and then apply the conversion factor between calories and joules.

🎯 Exam Tip: Ensure mass is in grams for calories and kilograms for joules when using specific heat values. Remember \(Q = mc\Delta T\) and that 1 calorie \(\approx\) 4.18 joules.

 

Question 5. When a substance having mass 2 kg absorbs 2000 cal of heat, its temperature increases by 10 °C. Find the specific heat of the substance.

Answer:0.1 cal/(g.°C)
In simple words: To find the specific heat, divide the heat absorbed by the product of the mass (in grams) and the temperature increase.

🎯 Exam Tip: Rearrange \(Q = mc\Delta T\) to \(c = \frac{Q}{m\Delta T}\). Be careful with units; convert kg to g if the expected answer is in cal/(g°C).

 

Question 6. Find the heat needed to raise the temperature of 100 g of a metal through 10 °C if the specific heat of the metal is 0.1 cal/g. °C.

Answer:100 cal
In simple words: Multiply the mass of the metal by its specific heat and the temperature change to determine the total heat required.

🎯 Exam Tip: Apply the heat transfer formula \(Q = mc\Delta T\). Ensure all units (mass in g, specific heat in cal/g°C, temperature in °C) are consistent for a direct calculation of heat in calories.

 

Question 7. If water of mass 80 g and temperature 40 °C is mixed with water of mass 20 g and temperature 25 °C, what will be the maximum temperature of the mixture?

Answer:37 °C
In simple words: The final mixture temperature is found by setting the heat lost by the hotter water equal to the heat gained by the colder water, considering their respective masses and initial temperatures.

🎯 Exam Tip: Use the principle of calorimetry: \(m_1 c \Delta T_1 = m_2 c \Delta T_2\). Since it's the same substance, 'c' cancels out. Solve for the final equilibrium temperature.

 

Question 8. A metal rod 2.5 m long, increases in length by 1.25 mm when it is heated from 10 °C to 60 °C. Find the coefficient of linear expansion of the metal.

Answer:1 x 10\(^{-5}\) °C\(^{-1}\)
In simple words: Calculate the coefficient of linear expansion by dividing the change in length by the product of the original length and the temperature difference, ensuring all length units are consistent.

🎯 Exam Tip: Remember to convert all length units to be consistent (e.g., mm to m). Apply the formula \(\lambda = \frac{\Delta L}{L_0 \Delta T}\) and pay attention to scientific notation in the final answer.

 

Question 9. The surface area of a metal plate is 2 × 10\(^{-2}\) m\(^2\) at 10 °C. When the plate is heated to 60 °C, its area increases by 0.2 m\(^2\). Find the coefficient of areal expansion of the metal.

Answer:2 x 10\(^{-5}\) °C\(^{-1}\)
In simple words: The coefficient of areal expansion is calculated by dividing the increase in surface area by the product of the original area and the temperature change.

🎯 Exam Tip: Ensure all area units are consistent (m²). The formula \(\sigma = \frac{\Delta A}{A_0 \Delta T}\) is essential. Double-check your exponent calculations for scientific notation.

 

Question 10. A metal ball has volume 50 cm\(^3\) at 0 °C and 50.4 cm\(^3\) at 100 °C. Find the volumetric expansion coefficient of the metal.

Answer:8 × 10\(^{-5}\) °C\(^{-1}\)
In simple words: Determine the volumetric expansion coefficient by dividing the change in volume by the product of the initial volume and the temperature difference.

🎯 Exam Tip: Use the formula \(\beta = \frac{\Delta V}{V_0 \Delta T}\). Keep volume units consistent (cm³) and temperature in °C for accurate calculation of the coefficient.

Ref. Project. Useful information: A bimetallic strip: A bimetallic strip consists of two strips of equal lengths but of different metals rivetted together. At room temperature the strip is straight.

 

Question 1. The principle on which a bimetallic strip works:

Answer:When two different metal strips of the same length at a given temperature are heated to the same higher temperature, they expand in different proportion. A bimetallic strip of brass and iron is straight at room temperature. The expansion of brass is nearly 1.5 times that of iron. Hence, when this bimetallic strip is heated, it bends, making the iron side concave.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक द्विधात्विक पट्टी (bimetallic strip) की कार्यप्रणाली को दर्शाता है। इसमें दो अलग-अलग धातुएँ (पीतल और लोहा) जुड़ी हुई हैं। जब पट्टी को गर्म किया जाता है, तो अलग-अलग धातुओं के भिन्न-भिन्न विस्तार के कारण यह मुड़ जाती है, जिसमें लोहा अंदर की ओर अवतल हो जाता है।
In simple words: A bimetallic strip works on the principle that different metals expand at different rates when heated, causing the strip to bend due to the unequal expansion of its two joined metal layers.

🎯 Exam Tip: The key concept of a bimetallic strip is the differential thermal expansion of two distinct metals. Be able to explain why it bends and which side becomes concave/convex.

 

Question 2. How a bimetallic strip is used in fire alarm:

Answer:A bimetallic strip of brass and iron is connected to a battery and an electric bell as shown in the diagram. One terminal of the bell is connected to a screw which is at a very small distance from the iron side of the strip. In case, there is an accidental fire, the bimetallic strip bends towards iron and touches the screw. Thus, the circuit is completed and the bell rings, thereby warning the people of the accidental fire.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र दिखाता है कि एक द्विधात्विक पट्टी (bimetallic strip) का उपयोग अग्नि अलार्म में कैसे किया जाता है। एक बैटरी और एक इलेक्ट्रिक घंटी एक द्विधात्विक पट्टी से जुड़े होते हैं। आग लगने पर, पट्टी गर्मी से मुड़ती है, लोहे की तरफ एक स्क्रू को छूती है, जिससे सर्किट पूरा हो जाता है और घंटी बजने लगती है।
In simple words: In a fire alarm, a bimetallic strip bends when heated by fire due to differential expansion, completing an electrical circuit that triggers an alarm bell.

🎯 Exam Tip: Understand the sequence: fire -> heat -> bimetallic strip bends -> circuit completion -> alarm. This is a practical application of thermal expansion for safety.