Maharashtra Board Class 10 Science Chapter 5 Heat Solutions

🎯 Exam Tip: Ensure a clear understanding of the formulas and units associated with latent heat and specific heat capacity, as this is a common matching question type.

Answer The Following Questions In One Sentence Each:

Question 1.State units of temperature.
Answer: Units of temperature: °C, °F and K (kelvin).
In simple words: Temperature is commonly measured in degrees Celsius, Fahrenheit, or Kelvin.

🎯 Exam Tip: Remember the three common temperature scales: Celsius, Fahrenheit, and Kelvin, with Kelvin being the SI unit.

Question 2.State units of energy.
Answer: Units of energy: the erg, the joule, the calorie.
In simple words: Energy can be measured in joules (SI unit), ergs, or calories.

🎯 Exam Tip: The joule is the standard international (SI) unit for energy, but calories and ergs are also commonly used, especially in older contexts or specific fields.

Question 3.State the relation between the joule and the calorie.
Answer: 1 calorie = 4.18 joules.
In simple words: One calorie is equivalent to approximately 4.18 joules of energy.

🎯 Exam Tip: This conversion factor (1 cal = 4.18 J) is crucial for converting between different energy units in numerical problems.

Question 4.State the relation between the erg and the joule.
Answer: 1 joule = 10⁷ ergs.
In simple words: One joule of energy is equal to ten million (10⁷) ergs.

🎯 Exam Tip: Understand the magnitude difference between SI units (joule) and CGS units (erg) for energy, specifically the 10⁷ factor.

Question 5.State the relation between the erg and the kilocalorie.
Answer: 1 kilocalorie = 4.18 × 10¹⁰ ergs.
In simple words: One kilocalorie is equivalent to 4.18 multiplied by 10 to the power of 10 ergs.

🎯 Exam Tip: For composite conversions, break them down: first kcal to cal, then cal to J, then J to erg. This reduces errors.

Question 6.State the relation between the joule and the kilocalorie.
Answer: 1 kilocalorie = 4.18 × 10³ joules.
In simple words: One kilocalorie equals 4180 joules.

🎯 Exam Tip: Remembering the direct conversion from kilocalorie to joule (4180 J) can save time in multi-step calculations.

Question 7.When heat energy is absorbed by an object, ΔT represents the rise in temperature. What would ΔT represent if the object loses heat energy?
Answer: If the object loses heat energy, ΔT would represent the decrease in temperature.
In simple words: ΔT signifies a change in temperature; if heat is absorbed, it's a rise, and if heat is lost, it's a fall.

🎯 Exam Tip: ΔT always means 'change in temperature', and its sign (positive for rise, negative for fall) corresponds to heat absorption or loss, respectively.

Answer The Following Questions:

Question 1.Define latent heat of fusion.
(OR)
What is latent heat of fusion? State its units.
Answer: When a solid is converted into liquid at constant temperature (melting point of the substance) the amount of heat absorbed by it is called the latent heat of fusion. Heat is a form of energy. Hence, latent heat is expressed in units joule, erg, calorie or kilocalorie.
In simple words: Latent heat of fusion is the energy absorbed by a solid to change into a liquid at its melting point without changing temperature, and its units are energy units.

🎯 Exam Tip: Always specify "at constant temperature" when defining latent heat to correctly convey that it's a phase change without temperature change.

Question 2.Define specific latent heat of fusion.
(OR)
What is specific latent heat of fusion? State its units.
Answer: The amount of heat energy absorbed at constant temperature by unit mass of a solid to convert into liquid phase is called the specific latent heat of fusion. It is expressed in units J/kg, erg/g, cal/g, kJ/kg and kcal/kg.
\[\text{Specific latent heat (L)} = \frac{\text{latent heat (Q)}}{\text{mass of the substance (m)}}\]

\( \implies \) Sl unit of specific latent heat = Sl unit of energy / Sl unit of mass = J/kg]
In simple words: Specific latent heat of fusion is the amount of heat energy needed to melt one unit of mass of a solid into a liquid at a constant temperature, and its units combine energy and mass.

🎯 Exam Tip: Differentiate between "latent heat" (total energy for any mass) and "specific latent heat" (energy per unit mass), ensuring you use the correct units (e.g., J/kg, cal/g).

Question 3.Explain the term latent heat of vaporization.
Answer: When a liquid is heated continuously, initially, its temperature increases. Later, at a certain stage, its temperature does not increase even when heat is supplied to it. At this temperature, heat absorbed by the liquid is used for breaking the bonds between its atoms or molecules, i.e., for doing work against the forces of attraction between the atoms or molecules and conversion into gaseous phase. This heat is called the latent heat of vaporization and the constant temperature at which this change of state occurs is called the boiling point of the liquid.
In simple words: Latent heat of vaporization is the energy absorbed by a liquid at its boiling point to transform into a gas without a temperature increase, as the heat breaks molecular bonds.

🎯 Exam Tip: Emphasize that latent heat of vaporization facilitates the phase transition by overcoming intermolecular forces, not by increasing kinetic energy (temperature).

Question 4.Define boiling point of a liquid.
(OR)
What is boiling point of a liquid?
Answer: The constant temperature at which a liquid transforms into gaseous state is called the boiling point of the liquid.
[Note: On application of pressure, the boiling point of a liquid is raised. On reducting the pressure, the boiling point is lowered.]
In simple words: The boiling point is the specific temperature at which a liquid rapidly changes into a gas.

🎯 Exam Tip: Remember that boiling point is pressure-dependent; higher pressure increases it, and lower pressure decreases it.

Question 5.Define specific latent heat of vaporization.
OR
What is specific latent heat of vaporization?
Answer: The amount of heat energy absorbed at constant temperature by unit mass of a liquid to convert into gaseous phase is called the specific latent heat of vaporization.
In simple words: Specific latent heat of vaporization is the heat needed to convert one unit of mass of a liquid into a gas at a constant temperature.

🎯 Exam Tip: Like fusion, specific latent heat of vaporization refers to a per-unit-mass quantity, indicating the energy required for the liquid-to-gas phase change.

Question 6.The specific latent heat of fusion of ice is 80 cal/g. Explain this statement.
Answer: When 1 g of ice at a pressure of one atmosphere and at a temperature 0 °C is converted into 1 g of water, heat absorbed by the ice is 80 cal.
In simple words: This statement means that 80 calories of heat energy are needed to change 1 gram of ice at 0°C into 1 gram of water at 0°C.

🎯 Exam Tip: When explaining a value, always include the mass (1g), initial and final states, and the constant temperature, along with the amount of heat absorbed/released.

Question 7.The specific latent heat of fusion of silver is 88.2 kJ/kg. Explain this statement.
Answer: When 1 kg of silver at a pressure of one atmosphere and at a temperature of 962 °C (melting point of silver) is converted into 1 kg of silver in liquid phase, heat absorbed by the silver is 88.2 kJ.
In simple words: This means 88.2 kilojoules of heat are required to melt 1 kilogram of solid silver into liquid silver at its melting point of 962°C.

🎯 Exam Tip: Be mindful of the units (kJ/kg) and the specific melting point of the substance when interpreting and explaining latent heat values for different materials.

Question 8.The specific latent heat of vaporization of water is 540 cal/g. Explain this statement.
Answer: When 1 g of water at a pressure of one atmosphere and at a temperature of 100 °C is converted into 1 g of steam, heat absorbed by the water is 540 cal.
In simple words: This statement means that 540 calories of heat energy are needed to convert 1 gram of water at 100°C into 1 gram of steam at the same temperature.

🎯 Exam Tip: Understand that this value highlights why steam causes more severe burns than boiling water; it carries a large amount of latent heat that is released upon condensation.

Question 9.Define regelation.
(OR)
What is regelation?
Answer: The phenomenon in which the ice converts to liquid due to applied pressure and then re-converts to ice once the pressure is removed is called regelation.
In simple words: Regelation is the process where ice melts under pressure and then refreezes once the pressure is released.

🎯 Exam Tip: Regelation explains phenomena like ice skating and the formation of snowballs, where pressure plays a crucial role in melting and refreezing.

Question 10.The terms hot and cold are used in relative context. Explain.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक प्रयोग दर्शाता है जिसमें गर्म और ठंडे की सापेक्षता समझाई गई है। इसमें तीन कटोरे (P, Q, R) हैं; P में ठंडा पानी, R में गर्म पानी और Q में गुनगुना पानी है। छात्र अपने हाथों को P और R में रखकर फिर Q में डुबोते हैं ताकि तापमान की सापेक्ष धारणा को समझा जा सके।
(1) Take three large bowls, P, Q and R. Fill bowl P with cold water, bowl Q with lukewarm water, and bowl R with hot water.
(2) Immerse your right hand in bowl P, and left hand in bowl R for about five seconds.
(3) Now, immerse both the hands in bowl Q at the same time.
(4) You will find that the water in bowl Q appears warm to your right hand, and cold to your left hand. Thus, the hand immersed in cold water for some time finds the lukewarm water hot while the one immersed in hot water finds the same lukewarm water cold. This experiment shows that the terms hot and cold are relative.
In simple words: Our perception of "hot" or "cold" is not absolute but depends on the previous temperature experience of our sensory organs, as demonstrated by comparing lukewarm water with hands previously in hot and cold water.

🎯 Exam Tip: Use the classic three-bowl experiment as an example to illustrate the subjective and relative nature of temperature perception, emphasizing that thermometers provide objective measurements.

Question 11.Draw a neat labelled diagram of Hope's apparatus. Explain how this apparatus can be used to demonstrate anomalous behaviour of water. Draw a graph of temperature of water against time.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र होप के उपकरण को दर्शाता है, जिसका उपयोग पानी के असामान्य व्यवहार को प्रदर्शित करने के लिए किया जाता है। इसमें पानी से भरा एक बेलनाकार पात्र है, जिसके मध्य भाग के चारों ओर बर्फ और नमक का मिश्रण रखा गया है। पानी के ऊपरी हिस्से (T2) और निचले हिस्से (T1) के तापमान को मापने के लिए दो थर्मामीटर लगे हुए हैं।
The figure shows Hope's apparatus. Initially, the cylindrical container in Hope's apparatus is filled with water at about 12 °C and the flat bowl is filled with a freezing mixture of ice and salt.
The temperature of water in the upper part of the container (T2) is recorded by thermometer T2 and that of water in the lower part of the container (T1) is recorded by thermometer T1. Figure shows variation of temperature of water with time.
Initially, both the thermometers show the same temperature (say, 12 °C). In a short time, the temperature shown by the lower thermometer starts decreasing, while the temperature shown by the upper thermometer does not change very much.
This process continues till the temperature shown by the lower thermometer falls to 4 °C and remains constant thereafter. This shows that in the temperature range 12 °C to 4 °C, the density of the water in the central part of the container goes on increasing and hence the water sinks to the bottom. It means that water contracts, i.e., its volume decreases as its temperature falls from 12 °C to 4 °C.
As the temperature of the water in the central part of the container becomes less than 4 °C, the temperature shown by the upper thermometer begins to fall rapidly to 0 °C. But the temperature shown by the lower thermometer remains constant (4 °C). Later, the heading shown by the lower thermometer decreases to 0 °C.
In the temperature range 4 °C to 0 °C, the water moves upward. This shows that the density of water goes on decreasing in this range. It means that water expands, i.e. its volume increases as its temperature falls from 4 °C to 0 °C.
Thus, the volume of a given mass of water is minimum at 4 °C, i.e., the density of water is maximum at 4 °C.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र होप के उपकरण में पानी के तापमान बनाम समय का ग्राफ दर्शाता है। इसमें दो वक्र (T1 और T2) हैं जो दिखाते हैं कि पानी का तापमान 4°C तक पहुंचने पर उसका घनत्व अधिकतम हो जाता है, और फिर 0°C तक ठंडा होने पर उसका आयतन फिर से बढ़ने लगता है, जो पानी के असामान्य व्यवहार को स्पष्ट करता है।
In the above figure, the point of intersection of the two curves shows the temperature at which the density of water is maximum. This temperature is 4 °C.
In simple words: Hope's apparatus demonstrates that water's density is highest at 4°C: as water cools, the colder, denser water (from 12°C to 4°C) sinks, then as it cools further (4°C to 0°C), it expands and less dense water rises, showing its anomalous behavior.

🎯 Exam Tip: When describing Hope's apparatus, clearly explain the role of both thermometers (T1 at bottom, T2 at top) and how their readings over time reveal water's maximum density at 4°C.

Question 12.A mountaineer climbing on the Everest, experienced the following facts. Explain each fact with the scientific reason: (1) He found j fishes alive below the ice (2) Time required for cooking was more as he went higher (3) He saw many times cliffs falling suddenly (4) He saw tubes carrying water broken.
Answer: Explanation:
(1) Water expands as its temperature decreases from 4 °C to 0 °C. Water is converted into ice at 0 °C. The density of water is more than that of ice. Fishes can remain alive in the water (at 4 °C) below the ice.
(2) At high altitudes, atmospheric pressure is low and hence water boils at a temperature lower than its normal boiling point. Therefore, the time required for cooking food is more at higher altitudes.
(3) Water expands while freezing. Hence, the water present in the crevices of the rocks exerts a tremendous pressure on the rocks, while freezing. Therefore, the cliffs fall.
(4) Water expands while freezing. Hence, the water in the tube exerts a large pressure on the tube, while freezing. Therefore, the tube carrying water breaks.
In simple words: These phenomena on Everest are due to water's anomalous expansion (allowing aquatic life, breaking rocks/pipes) and the effect of reduced atmospheric pressure at high altitudes (lowering water's boiling point, thus increasing cooking time).

🎯 Exam Tip: This question is a good test of applying multiple concepts: anomalous expansion of water, effect of pressure on boiling point, and density differences between ice and water.

Question 13.What is humidity?
Answer: The moisture, i.e., the presence of water vapour, in the atmosphere is called humidity.
In simple words: Humidity refers to the amount of water vapor present in the air.

🎯 Exam Tip: Keep the definition of humidity concise and focus on water vapor as the key component.

Question 14.When is air said to be saturated with water vapour?
Answer: When air contains maximum possible water vapour, it is said to be saturated with water vapour at that temperature.
In simple words: Air is saturated with water vapor when it cannot hold any more moisture at its current temperature.

🎯 Exam Tip: The concept of saturation is crucial for understanding dew point and relative humidity; always mention "at that temperature" as capacity changes with temperature.

Question 15.What does the amount of water vapour needed to saturate air depend on?
Answer: The extent of water vapour needed to saturate air depends on the temperature. The greater the temperature, the greater is the amount of water vapour needed to saturate air.
In simple words: The amount of water vapor needed to saturate air depends directly on its temperature; warmer air can hold more moisture.

🎯 Exam Tip: Remember the direct relationship between air temperature and its capacity to hold water vapor; this is fundamental to atmospheric science.

Question 16. What is air said to be unsaturated with water vapour?
Answer: When air contains water vapour less than its capacity to hold water vapour at that temperature, it is said to be unsaturated with water vapour.
In simple words: Air is unsaturated when it holds less water vapor than it possibly could at its current temperature, meaning it can still absorb more moisture.

🎯 Exam Tip: Understanding the concepts of saturation and unsaturation is crucial for explaining weather phenomena like cloud formation and dew.

Question 17. What is dew point temperature?
(OR)
Define dew point temperature.
Answer: If the temperature of unsaturated air is decreased, a temperature is reached at which the air becomes saturated with water vapour. This temperature is called the dew point temperature.
In simple words: Dew point temperature is the specific temperature at which air becomes completely saturated with water vapor, leading to condensation if cooled further.

🎯 Exam Tip: The dew point is a key indicator of atmospheric moisture and is often tested in questions related to condensation.

Question 18. Name the physical quantity used to express the amount of water vapour present in air.
Answer: Absolute humidity.
In simple words: Absolute humidity is the measure of the actual amount of water vapor in a given volume of air.

🎯 Exam Tip: Differentiate between absolute and relative humidity; questions often test this distinction.

Question 19. Define absolute humidity.
(OR)
What is absolute humidity? State its unit.
Answer: The mass of water vapour present in a unit volume of air is called absolute humidity. Generally it is expressed in kg/m³.
In simple words: Absolute humidity quantifies the mass of water vapor per unit volume of air, usually measured in kilograms per cubic meter.

🎯 Exam Tip: Always remember to state the correct units for physical quantities like absolute humidity.

Question 20. Define relative humidity.
(OR)
What is relative humidity? Write the formula for % relative humidity.
Answer: The ratio of the actual mass of water vapour content in the air for a given volume and temperature to that required to make the same volume of air saturated with water vapour at the same temperature is called the relative humidity.
% Relative humidity = [the actual mass of water vapour content in the air for a given volume and temperature \( \div \) the mass of water vapour required to make the same volume of air saturated with water vapour at the same temperature] \( \times \) 100%.
In simple words: Relative humidity indicates how much water vapor is currently in the air compared to the maximum it can hold at that temperature, expressed as a percentage.

🎯 Exam Tip: The formula for relative humidity is important for numerical problems; ensure you understand each component.

Question 21. What is the value of relative humidity at the dew point temperature?
Answer: At the dew point temperature, relative humidity is 100%.
In simple words: When the air reaches its dew point, it is fully saturated with moisture, making the relative humidity 100%.

🎯 Exam Tip: Remember that 100% relative humidity at the dew point is a critical concept for understanding condensation.

Question 22. The mass of water vapour in air enclosed in a certain space is 60 g and the mass of water vapour needed to saturate the same air with water vapour under the same conditions is 100 g. What is the corresponding % relative humidity?
Answer: Here, % relative humidity = \( \left(\frac{60 \text{g}}{100 \text{g}}\right) \times 100\% = 60\% \)
In simple words: To find the relative humidity, divide the actual water vapor mass (60g) by the saturation capacity (100g) and multiply by 100%, which gives 60%.

🎯 Exam Tip: Practice similar numerical problems using the relative humidity formula to master its application.

Question 23. During winter, sometimes we see a white trail at the back of a flying aeroplane in a clear sky. Explain why.
Answer: In winter, air temperature is low. Hence, when an aeroplane flies, the vapour released by its engine condenses and forms white clouds. If the relative humidity of the air surrounding the plane is high, we see this white trail at the back of the plane for a long time before it disappears. If the relative humidity is low, the white trail is short and disappears quickly. If the relative humidity is very low, there is no formation of the white trail.
In simple words: Aeroplane contrails form because hot engine exhaust, containing water vapor, rapidly cools and condenses into visible ice crystals or water droplets in the cold, often humid, upper atmosphere.

🎯 Exam Tip: This question links humidity, condensation, and atmospheric conditions; focus on the role of temperature and relative humidity in vapor condensation.

Question 24. State two effects of humidity present in atmosphere.
Answer: Effects of humidity present in atmosphere: When the temperature of air falls below the dew point, dew and fog are formed.
In simple words: Humidity influences the formation of dew and fog when air temperature drops to or below the dew point, causing water vapor to condense.

🎯 Exam Tip: Be prepared to explain how humidity impacts various weather phenomena, not just dew and fog.

Question 25. Explain how dew and fog are formed.
(OR)
Write a short note on formation of dew and fog.
Answer: At a particular temperature, a given volume of air can contain a certain maximum amount of water vapour. Normally, the temperature of air during the day is such that air is not saturated with water vapour present in it.
As the temperature falls, the capacity of air to hold water vapour becomes less.
During a cold night, the temperature of air may fall to the dew point, or even below the dew point. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed. If the water vapour condenses on the fine dust particles present in the atmosphere, mist or fog is formed.
In simple words: Dew forms when water vapor condenses on cold surfaces below the dew point, while fog arises when water vapor condenses on atmospheric dust particles, creating a visible cloud near the ground.

🎯 Exam Tip: Clearly distinguish between dew and fog formation by noting where the condensation occurs-on surfaces for dew and on dust particles in the air for fog.

Question 26. State the units of heat.
Answer: Units of heat: joule, erg, calorie, kilocalorie.
In simple words: Heat, a form of energy, is commonly measured in units such as joules, ergs, calories, or kilocalories.

🎯 Exam Tip: Familiarize yourself with all common units of heat and their conversions for numerical problems.

Question 27. Define the kilocalorie.
Answer: The amount of heat necessary to raise the temperature of 1 kg of water by 1 °C from 14.5 °C to 15.5 °C is called one kilocalorie.
In simple words: A kilocalorie is defined as the energy needed to increase the temperature of 1 kilogram of water by 1 degree Celsius, specifically between 14.5 °C and 15.5 °C.

🎯 Exam Tip: Note the precise temperature range (14.5 °C to 15.5 °C) when defining a kilocalorie, as it is a standard reference.

Question 28. Define the calorie.
Answer: The amount of heat necessary to raise the temperature of 1 g of water by 1 °C from 14.5 °C to 15.5 °C is called one calorie.
In simple words: A calorie is the amount of heat required to raise the temperature of 1 gram of water by 1 degree Celsius, specifically from 14.5 °C to 15.5 °C.

🎯 Exam Tip: Pay attention to the mass (1g vs 1kg) when differentiating between calorie and kilocalorie definitions.

Question 29. State the relation between the kilocalorie and the calorie.
Answer: 1 kilocalorie = 10³ calories.
In simple words: One kilocalorie is equivalent to one thousand calories, making it a larger unit for measuring heat energy.

🎯 Exam Tip: This conversion factor is fundamental and frequently used in energy calculations.

Question 30. Study the following procedure and answer the questions below: (Practice Activity Sheet - 2)
1. Take 3 spheres of iron, copper and lead of equal mass.
2. Put all the 3 spheres in boiling water in a beaker for some time.
3. Take 3 spheres out of the water. Put them immediately on a thick slab of wax.
4. Note the depth that each sphere goes into the wax.
(i) Which property of a substance can be studied with this procedure?
(ii) Describe that property in minimum words.
(iii) Explain the rule of heat exchange with this property.
Answer: (i) Specific heat.
(ii) Specific heat: The amount of heat energy required to raise the temperature of a unit mass of an object by 1 °C.
(iii) According to the rule/principle of heat exchange, heat energy lost by the hot object = heat energy gained by the cold object.
In this activity, heat absorbed by the iron sphere is transmitted more in the wax, hence the sphere goes deepest into the wax, while the lead sphere absorbs less heat, resulting in less transmission of heat in the wax, hence, the sphere goes the least depth into the wax.
In simple words: This experiment demonstrates specific heat capacity; iron, having a higher specific heat, retains more heat and melts deeper into the wax than lead, which has a lower specific heat.

🎯 Exam Tip: When explaining experiments, clearly link the observed phenomena to the underlying scientific principle, in this case, specific heat and heat exchange.

Question 31. Write the symbol for specific heat capacity. State the units of specific heat capacity.
Answer: Symbol for specific heat capacity: c.
Units of specific heat capacity: J/kg·°C,
erg/g·°C, cal/g·°C, kcal/kg·°C.
[ Notes: (1) Specific heat capacity \( = \frac{\text{heat absorbed by an object}}{\text{mass of the object } \times \text{ rise in temperature of the object}} \)
\( \therefore \) Unit of specific heat capacity \( = \frac{\text{unit of heat}}{\text{unit of mass } \times \text{ unit of temperature}} \)
In SI, heat is expressed in joule (J), mass in kg and temperature in kelvin(K).
\( \therefore \) SI unit of specific heat capacity \( = \frac{\text{J}}{\text{kg}\cdot\text{K}} \). (2) The specific heat capacity of a substance depends upon its constituent particles (atoms, molecules, etc.), interaction between them, structure of the substance (atomic/molecular arrangement), temperature of the substance, etc.]
In simple words: The symbol for specific heat capacity is 'c', and its common units include Joules per kilogram per degree Celsius (J/kg·°C) or calories per gram per degree Celsius (cal/g·°C).

🎯 Exam Tip: Always remember the symbol 'c' for specific heat capacity and be able to list its various units, especially the SI unit.

Question 32. Explain the principle of heat exchange. Ans. Suppose two objects A and B at different temperature T₁ and T2 respectively are enclosed in a box of heat resistant material as shown in figure.
Answer:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक ऊष्मा प्रतिरोधी बॉक्स के अंदर दो वस्तुओं, A (गर्म वस्तु) और B (ठंडी वस्तु) को दर्शाता है। दोनों वस्तुएं एक-दूसरे के संपर्क में हैं, जिससे उनके बीच ऊष्मा का आदान-प्रदान हो रहा है, जबकि बॉक्स उन्हें बाहरी वातावरण से अलग रखता है। यह व्यवस्था ऊष्मा के आदान-प्रदान के सिद्धांत को समझने में मदद करती है।
Let m₁ = mass of A, m2 = mass of B, c₁ = specific heat capacity of A, c2 = specific heat capacity of B and T = common temperature attained by A and B after the heat exchange between A and B. Here, no heat leaves the box or enters the box from outside. Hence, if T₁ > T2, heat energy lost by A (Q1) = heat energy gained by B (Q2).
\( \therefore \) m₁c₁ (T1 - T) = m2c2 (T - T2)
[Note: If m₁, C1, T1, T, m2 and T2 are known, c2 can be determined.]
In simple words: The principle of heat exchange states that in an isolated system, the total heat lost by hotter objects equals the total heat gained by colder objects until thermal equilibrium is reached.

🎯 Exam Tip: When explaining the principle of heat exchange, ensure you mention an isolated system and the concept of thermal equilibrium.

Question 33. The specific heat capacity of silver is 0.056 kcal/kg·°C. Explain this statement.
Answer: The amount of heat needed to raise the temperature of 1 kg of silver by 1 °C is 0.056 kcal.
In simple words: This statement means that 0.056 kilocalories of heat energy are required to increase the temperature of 1 kilogram of silver by 1 degree Celsius.

🎯 Exam Tip: When interpreting specific heat capacity values, always specify the mass and temperature change involved for clarity.

Question 34. Explain how the specific heat capacity of a solid can be determined (measured) by the method of mixture.
Answer: A hot solid is put in water in a calorimeter. The mixture is stirred continuously and the maximum temperature of the mixture is measured with a thermometer. Heat exchange between the hot solid, water and calorimeter results in sill bodies attaining the same temperature after some time. Hence, according to the principle of heat exchange, heat lost by the solid = heat gained by the water in the calorimeter + heat gained by the calorimeter.
Now, heat lost by the solid (Q) = mass of the solid \( \times \) its specific heat capacity \( \times \) decrease in its temperature, heat gained by the water (Q1) = mass of the water \( \times \) its specific heat capacity \( \times \) increase in its temperature and heat gained by the calorimeter (Q2) = mass of the calorimeter \( \times \) its specific heat capacity \( \times \) increase in its temperature.
Heat lost by the hot object = heat gained by the calorimeter + heat gained by the water. Q = Q2 + Q1
Using this equation, the specific heat capacity of the solid can be determined (measured) when the other quantities are known.
In simple words: The method of mixtures determines a solid's specific heat by measuring the heat exchanged when a hot solid is placed in a calorimeter with water, applying the principle that heat lost equals heat gained.

🎯 Exam Tip: Remember the core equation (Heat lost = Heat gained) and list all components (solid, water, calorimeter) when describing the method of mixtures.

Question 1. Even though heat is supplied to boiling water, there is no increase in its temperature.
Answer: Once water starts boiling, all the heat supplied to it is used in conversion of water into steam at the boiling point of water. Hence, there is no rise in its temperature.
In simple words: When water boils, the added heat energy is used for the phase change (liquid to gas) rather than increasing the temperature, which remains constant at the boiling point.

🎯 Exam Tip: This phenomenon highlights the concept of latent heat, where energy is used for phase transitions instead of temperature change.

Question 2. Burns from steam are worse those from boiling water at the same temperature.
Answer: 1. A given quantity of steam contains more heat than the same quantity of boiling water at the same temperature.
2. When steam comes in contact with one's body, it releases extra heat of 540 calories per gram and causes a more serious burn than that caused by boiling water.
In simple words: Steam at 100 °C causes more severe burns than boiling water at 100 °C because steam releases additional latent heat of vaporization (540 cal/g) upon condensing on the skin, causing greater energy transfer.

🎯 Exam Tip: Focus on the latent heat of vaporization that steam releases upon condensation; this additional energy is the key reason for more severe burns.

Question 3. In winter, the pipelines carrying water burst in cold countries.
Answer: 1. In cold countries, in winter, the temperature of the atmosphere falls below 0 °C. When the temperature of water falls below 4 °C, it expands. Even if ice is formed, there is an increase in the volume.
2. As there is no room for expansion, water (or ice) exerts a large pressure on the pipes. Hence, the pipelines carrying water burst.
In simple words: Water expands anomalously when it cools below 4 °C and freezes, increasing its volume and exerting immense pressure on rigid pipes, causing them to burst.

🎯 Exam Tip: This question is a classic application of the anomalous behavior of water; specifically, its expansion upon freezing.

Question 4. If crushed ice is pressed and then the pressure is released, a lump of ice is formed.
Answer: 1. When crushed ice is pressed, its melting point is lowered and some ice melts to form water.
2. When pressure is released, the melting point becomes normal and the water freezes to form ice forming a lump.
In simple words: Pressing crushed ice lowers its melting point, causing some ice to melt; when pressure is released, the melting point returns to normal, and the water refreezes, forming a solid lump.

🎯 Exam Tip: This is an example of regelation; remember the relationship between pressure, melting point, and refreezing.

Question 5. In cold countries, in winter, even when the water of lakes freezes, aquatic animals and plants can survive.
Answer: 1. In cold countries, in winter, a layer of ice is formed on the surface of lakes when the atmospheric temperature falls below 0 °C. However, below this layer, there is water at 4 °C.
2. Ice, being a bad conductor of heat, does not allow transfer of heat from this water to the atmosphere. Hence, aquatic animals and plants can survive in this water.
In simple words: Aquatic life survives in frozen lakes due to water's anomalous expansion; water at 4°C, being densest, sinks to the bottom, and the insulating ice layer prevents further cooling.

🎯 Exam Tip: This question also relates to the anomalous behavior of water, emphasizing the importance of the 4 °C maximum density for aquatic life.

Question 6. Water droplets are seen on the outer surface of a cold drink bottle.
Answer: 1. The temperature of the outer surface of a cold drink bottle is less than that of the atmosphere.
2. Therefore, the excess of water vapour from the air condenses to form droplets on the outer surface of the cold drink bottle.
In simple words: Water droplets form on a cold bottle because the bottle's surface cools the surrounding air below its dew point, causing water vapor in the air to condense onto the cold surface.

🎯 Exam Tip: This is a simple yet effective example of condensation and dew point; ensure you clearly explain the role of temperature difference.

Question 7. During cold nights, sometimes dew is formed.
Answer: 1. During a cold night, the temperature of air may fall to the dew point, or even below the dew point. 2. If the temperature falls below the dew point, the excess of water vapour in air condenses on the surfaces of cold bodies and dew is formed.
In simple words: Dew forms on cold nights when surfaces cool the nearby air to its dew point, causing the water vapor in that air to condense into liquid droplets.

🎯 Exam Tip: Relate dew formation directly to the air temperature dropping to the dew point, leading to condensation on surfaces.

Question 8. When you enter a warm room after being outside on a frosty early morning, your spectacles 'steam up'.
Answer: 1. On a frosty early morning, the temperature of air outside a warm room is lower than the dew point.
2. Hence, when you enter the room from outside, some water vapour in the room condenses on the glass of your spectacles, i.e., the spectacles 'steam up'.
In simple words: Spectacles 'steam up' when moving from cold to warm air because the cold lenses cool the warm, humid room air below its dew point, causing water vapor to condense on them.

🎯 Exam Tip: This is another example of condensation, where the cold surface (spectacles) causes warm, humid air to reach its dew point.

Question 9. A plastic bottle, completely filled with water, when kept in a freezer, is likely to break.
Answer: The temperature of air in the freezer (deep freeze) compartment of a refrigerator is less than 0 °C. 2. When a plastic bottle, completely filled with water, is kept in this compartment, the temperature of water falls below 4 °C and the water expands. Even when water freezes and ice is formed, there is an increase in the volume. It exerts a large pressure on the sides of the bottle and hence the bottle is likely to break.
In simple words: A full plastic bottle in a freezer breaks because water expands as it cools below 4 °C and freezes into ice, increasing its volume and creating immense pressure within the confined bottle.

🎯 Exam Tip: Similar to pipelines bursting, this question also relies on the anomalous expansion of water upon freezing.

Question 10. The outer surface of a beaker containing ice cubes becomes wet in a short while.
Answer: 1. When ice cubes are placed in a beaker, ice starts melting. The heat required for melting is absorbed from the surrounding air and also from the beaker to some extent.
Hence, the temperature of the air and beaker falls.
2. The capacity of air to hold water vapour depends upon the temperature of the air, and this capacity decreases as the temperature decreases. At a certain low temperature, the surrounding air becomes saturated with water vapour present in it. As the temperature falls further, the air is unable to hold all the water vapour.
Hence, the extra water vapour starts condensing on the cold outer surface of the beaker in the form of minute drops. Therefore, the outer surface of the beaker containing ice cubes becomes wet in a short while.
In simple words: The beaker's surface becomes wet as ice cubes cool it, causing the surrounding humid air to reach its dew point and condense its excess water vapor into tiny droplets on the cold glass.

🎯 Exam Tip: This explanation combines heat absorption (melting ice), temperature drop, and the concept of dew point for condensation.

Distinguish between the following:

Question 1. Absolute humidity and Relative humidity.
Answer: Absolute humidity:
1. Absolute humidity is the mass of water vapour present in a unit volume of air.
2. It is commonly expressed in kg/m³.
Relative humidity:
1. Relative humidity is the ratio of the mass of water vapour in a given volume of air at a given temperature to the mass of water vapour required to saturate the same volume of air at the same temperature.
2. It does not have unit.
In simple words: Absolute humidity measures the actual amount of water vapor in air (mass per volume, e.g., kg/m³), while relative humidity expresses this amount as a percentage of the maximum water vapor the air can hold at that temperature, hence unitless.

🎯 Exam Tip: When distinguishing, always mention the definition, units (or lack thereof), and how each relates to the total moisture content and saturation level.

Solve the following examples/Numerical problems:
[Use the data given in the Tables on pages 130 and 131.]

Problem 1. Calculate the amount of heat required to convert 5 g of ice of 0 °C into water at 0 °C. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: Here, m = 5 g, L = 80 cal/g; Q = ?
Amount of heat required, Q = mL
= 5 g \( \times \) 80 cal/g
= 400 calories.
In simple words: To melt 5 grams of ice at 0°C into water at 0°C, multiply its mass by the specific latent heat of fusion (80 cal/g) to find the total heat required, which is 400 calories.

🎯 Exam Tip: For phase change problems at constant temperature, use Q = mL, where L is the latent heat of fusion or vaporization.

Problem 2. Find the amount of heat required to convert 10 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 10 g \( \times \) 540 cal/g
= 5400 calories.
In simple words: To convert 10 grams of boiling water into steam, multiply its mass by the latent heat of vaporization (540 cal/g) to get 5400 calories of heat.

🎯 Exam Tip: Ensure you use the correct latent heat (fusion or vaporization) based on the phase change described in the problem.

Problem 3. Calculate the amount of heat required to convert 15 g of water at 100 °C into steam. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: m = 15 g, L = 540 cal/g; Q = ?
Amount of heat required, Q = mL
= 15 g \( \times \) 540 cal/g
= 8100 calories.
In simple words: To convert 15 grams of water at 100°C to steam, multiply its mass by the latent heat of vaporization (540 cal/g), resulting in 8100 calories needed.

🎯 Exam Tip: This problem reinforces the direct application of the latent heat formula for vaporization.

Problem 4. How many calories of heat will be absorbed when 3 kg of ice at 0 °C melts?
Solution: m = 3 kg = 3000 g; L = 80 cal/g; Q = ?
Quantity of heat absorbed, Q = mL
= 3000 g \( \times \) 80 cal/g
\( \therefore \) Q = 240000 calories.
In simple words: To melt 3 kg (3000g) of ice at 0°C, multiply its mass by the latent heat of fusion (80 cal/g), requiring 240,000 calories.

🎯 Exam Tip: Always convert mass to the appropriate unit (grams in this case) to match the units of latent heat.

Problem 5. Calculate the amount of heat required to convert 10 g of water at 30 °C into steam at 100 °C. (Specific latent heat of vaporization of water = 540 cal/g)
Solution: Here, m = 10 g; c = 1 cal/g·°C
T2-T1 = 100 °C - 30 °C = 70 °C; L = 540 cal/g; Q = ?
Amount of heat required, Q = mc (T2 - T₁) + mL
= 10 g \( \times \) 1 cal/g·°C \( \times \) 70 °C + 10 g \( \times \) 540 cal/g
= 700 cal + 5400 cal
\( \therefore \) Q = 6100 calories.
In simple words: To convert 10g of water from 30°C to steam at 100°C, first calculate the heat needed to raise its temperature to 100°C, then add the latent heat required for vaporization.

🎯 Exam Tip: For problems involving both temperature change and phase change, calculate heat absorbed/released for each stage separately and then sum them up.

Problem 6. If water of mass 80 g and temperature 45 °C is mixed with water of mass 20 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution: Data: m₁ = 80 g, T₁ = 45 °C, m2 = 20 g,
T2 = 30 °C, T = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
\( \therefore \) m₁c (T1 - T) = m2c (T - T2)
\( \therefore \) m₁T₁ - m₁T = m2T - m2T2
\( \therefore \) m₁T1 + m2T2 = (m1 + m2)T
\( \therefore \) Maximum temperature of the mixture,
T = \( \frac{m_1T_1 + m_2T_2}{m_1 + m_2} \)
= \( \frac{80 \text{g} \times 45 \text{°C} + 20 \text{g} \times 30 \text{°C}}{80 \text{g} + 20 \text{g}} \)
= \( \left(\frac{80 \times 45 + 20 \times 30}{100}\right) \text{°C} \)
= (36 + 6) °C
= 42°C.
In simple words: To find the final temperature when two water masses at different temperatures mix, apply the principle of heat exchange, where the heat lost by the hotter water equals the heat gained by the colder water.

🎯 Exam Tip: Remember that for mixing problems with the same substance, specific heat capacity 'c' often cancels out, simplifying the calculation.

Problem 7. When water of mass 70 g and temperature 50 °C is added to water of mass 30 g, the maximum temperature of the mixture is found to be 41 °C. Find the temperature of water of mass 30 g before hot water was added to it.
Solution: Data: m₁ = 70 g, T₁ = 50 °C, m2 = 30 g, T = 41 °C, T2 = ?
According to the principle of heat exchange, heat lost by hot water = heat gained by cold water
\( \therefore \) m₁c (T1 - T) = m2c (T - T2)
\( \therefore \) m₁T₁ - m₁T = m2T - m2T2
\( \therefore \) m2T2 = (m1 + m2) T - m₁T1
\( \therefore \) T2 = \( \frac{(m_1+m_2) T - m_1T_1}{m_2} \)
= \( \frac{(70 \text{g} + 30 \text{g}) 41 \text{°C} - 70 \text{g} \times 50 \text{°C}}{30 \text{g}} \)
= \( \left(\frac{100 \times 41 - 70 \times 50}{30}\right) \text{°C} \)
= \( \left(\frac{4100 - 3500}{30}\right) \text{°C} \)
= \( \frac{600}{30} \) °C
= 20 °C
This is the required temperature.
In simple words: Using the principle of heat exchange and knowing the final mixture temperature, we can rearrange the equation to solve for the initial unknown temperature of the colder water.

🎯 Exam Tip: Be careful with algebraic rearrangement when solving for an unknown initial temperature; double-check your substitutions and calculations.

Problem 8. Find the heat needed to raise the temperature of a silver container of mass 100 g by 10 °C. (c = 0.056 cal/g·°C)
Solution: Data: m = 100 g, \( \Delta \)T = 10 °C, c = 0.056 cal/g·°C
Heat needed to raise the temperature of the container = mc \( \Delta \)T
= 100 g \( \times \) 0.056 cal/g·°C \( \times \) 10 °C.
= 56 calories.
In simple words: To calculate the heat needed, multiply the silver container's mass (100g) by its specific heat capacity (0.056 cal/g·°C) and the desired temperature change (10°C), resulting in 56 calories.

🎯 Exam Tip: This is a direct application of the specific heat formula Q = mc\( \Delta \)T; ensure units are consistent.

Problem 9. If steam of mass 100 g and temperature 100 °C is released on an ice slab of temperature 0 °C, how much ice will melt?
Solution: Data: m₁ = 100 g, L₁ = 540 cal/g,
T₁ = 100 °C, mass of ice, m = ?, L2 = 80 cal/g, c (water) = 1 cal/g·°C
According to the principle of heat exchange, heat lost by hot body = heat gained by cold body. Conversion of steam into water:
Q₁ = m₁L₁ = 100 g \( \times \) 540 cal/g = 54000 cal
Decrease in the temperature of this water to 0 °C:
Q2 = m₁c \( \times \) (T₁ - 0 °C) = 100 g \( \times \) 1 cal/g·°C \( \times \) (100 °C - 0 °C) = 10000 Cal
Melting of ice: Q3 = mL2
= m \( \times \) 80 cal/g
Now, Q1 + Q2 = Q3
\( \therefore \) (54000 + 10000) cal = m \( \times \) 80 cal/g
\( \therefore \) m = \( \frac{64000}{80} \) = 800 g
800 g of ice will melt.
In simple words: When steam condenses and cools on ice, the total heat released (from steam condensation and cooling water) is absorbed by the ice to melt it; we equate these energies to find the melted ice mass.

🎯 Exam Tip: Break down complex problems involving multiple phase changes and temperature changes into individual heat transfer steps, then apply the principle of heat exchange.

Numerical problems for practice:

Problem 1. Calculate the amount of heat required to convert 80 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: 6400 cal
In simple words: To convert 80g of ice to water at 0°C, multiply its mass by the latent heat of fusion (80 cal/g), resulting in 6400 calories.

🎯 Exam Tip: This is a direct application of the latent heat of fusion formula; ensure correct units and calculation.

Problem 2. Find the heat required to convert 20 g of ice at 0 °C into water at the same temperature. (Specific latent heat of fusion of ice = 80 cal/g)
Solution: 1600 cal
In simple words: To melt 20g of ice at 0°C into water at 0°C, multiply its mass by the latent heat of fusion (80 cal/g), yielding 1600 calories.

🎯 Exam Tip: Remember Q=mL for phase changes without temperature change, ensuring correct mass and latent heat values.

Problem 3. Calculate the quantity of heat released during the conversion of 10 g of ice cold water (temperature 0 °C) into ice at the same temperature. (Specific latent heat of freezing of water = 80 cal/g)
Solution: 800 cal
In simple words: To calculate the heat released when 10g of water freezes at 0°C, multiply its mass by the latent heat of freezing (80 cal/g), which gives 800 calories.

🎯 Exam Tip: Latent heat of freezing is numerically equal to the latent heat of fusion; it's simply heat released instead of absorbed.

Problem 4. How many calories of heat will be absorbed when 2 kg of ice at 0 °C melts? (Specific latent heat of fusion of ice = 80 cal/g)
Solution: 160000 cal
In simple words: To melt 2 kg (2000g) of ice at 0°C, multiply its mass by the latent heat of fusion (80 cal/g), which requires 160,000 calories.

🎯 Exam Tip: Pay close attention to unit conversions (kg to g) to match the given specific latent heat units.

Problem 5. How much heat will be required to convert 20 g of water at 100 °C into steam at 100 °C? (Specific latent heat of vaporization of water = 540 cal/g)
Solution: 10800 cal
In simple words: To convert 20g of water at 100°C to steam at the same temperature, multiply its mass by the latent heat of vaporization (540 cal/g), requiring 10,800 calories.

🎯 Exam Tip: This is a straightforward application of the latent heat of vaporization formula; ensure accurate multiplication.

Problem 6. Find the heat absorbed by 25 g of water at 100 °C to convert into steam at the same temperature. (Specific latent heat of vaporization of water = 540 cal/g.)
Solution: 13500 cal
In simple words: To convert 25g of water at 100°C to steam at the same temperature, multiply its mass by the latent heat of vaporization (540 cal/g), resulting in 13,500 calories of heat absorbed.

🎯 Exam Tip: Identify the correct latent heat value for vaporization (540 cal/g) and apply the Q=mL formula.

Problem 7. If water of mass 60 g and temperature 50 °C is mixed with water of mass 40 g and temperature 30 °C, what will be the maximum temperature of the mixture?
Solution: 42 °C
In simple words: When 60g of water at 50°C mixes with 40g of water at 30°C, the final mixture temperature, found by applying the principle of heat exchange, will be 42°C.

🎯 Exam Tip: For mixing problems involving only temperature changes of the same substance, the specific heat cancels out, simplifying calculations to a weighted average of temperatures.

Problem 8. If water of mass 60 g and temperature 60 °C is mixed with water of mass 60 g and temperature 40 °C, what will be the maximum temperature of the mixture?
Solution: 50 °C
In simple words: When equal masses of water at 60°C and 40°C are mixed, the final temperature will be the average of the two, which is 50°C.

🎯 Exam Tip: If equal masses of the same substance at different temperatures are mixed, the final temperature is simply the average of the initial temperatures.

Problem 9. Find the heat needed to raise the temperature of a piece of iron of mass 500 g by 20 °C. (c = 0.110 cal/g·°C)
Solution: 1100 cal
In simple words: To raise the temperature of 500g of iron by 20°C, multiply its mass, specific heat capacity (0.110 cal/g·°C), and the temperature change, yielding 1100 calories.

🎯 Exam Tip: Use the formula Q = mc\( \Delta \)T, ensuring that the mass is in grams to match the specific heat capacity units.