GSEB Class 11 Physics Solutions Chapter 11 Thermal Properties of Matter

 

Question 1. The triple point of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperature on the Celsius and Fahrenheit scales?
Answer: Here, the triple point of neon, \( T_{1} = 24.57 K \), and the triple point of \( CO_{2} \), \( T_{2} = 216.55 K \). We understand that the temperature of a body on Celsius, Fahrenheit, Rankine, and absolute scales are connected as:
\[ \frac{C-0}{100-0} = \frac{F-32}{212-32} = \frac{R-0}{80-0} = \frac{T-273.15}{100} \]
On the Celsius scale, we use the connection:
\[ \frac{C-0}{100-0} = \frac{T-273.15}{100} \]
or \( C = T-273.15 \)
For neon, \( t_{1}^{\circ}{C} = 24.57 - 273.15 = -248.58^{\circ}C \)
For \( CO_{2} \), \( t_{2}^{\circ}{C} = 216.55 - 273.15 = -56.6^{\circ}C \)
On the Fahrenheit Scale:
\[ \frac{F-32}{180} = \frac{T-273.15}{100} \]
For neon \( F_{1} = (T_{1} - 273.15) \times \frac{9}{5} + 32 \)
\( = (24.57 - 273.15) \times \frac{9}{5} + 32 \)
\( = -248.58 \times \frac{9}{5} + 32 \)
\( = -415.44^{\circ}F \)
For \( CO_{2} \),
\( F_{2} = (T_{2}-273.15) \times \frac{9}{5} + 32 \)
\( = (216.55 - 273.15) \times \frac{9}{5} + 32 \)
\( = -56.6 \times \frac{9}{5} + 32 \)
\( = -69.88^{\circ}F \).
In simple words: To convert Kelvin temperatures to Celsius, subtract 273.15. To convert to Fahrenheit, use the formula involving (T-273.15) multiplied by 9/5 and add 32.

Exam Tip: Remember the precise conversion factors for temperature scales: 273.15 for Kelvin to Celsius, and the (9/5) factor with 32 for Fahrenheit. Pay close attention to negative signs.

 

Question 2. Two absolute scales A and B have triple-points of water defined to be 200 A and 350 B. What is the relation between \( T_{A} \) and \( T_{B} \)?
Answer: Here, the triple point of water on absolute scale A is 200 A, and the triple point of water on absolute scale B is 350 B. We also know that the triple point of water on the absolute scale is \( 273.16 K \). This means that temperature 200 A and 350 B on absolute scale A are equal to temperature 273.16 on absolute scale. The absolute scales, therefore, measure the triple point as 200A and 350B.
So, the size of one degree of Kelvin scale on absolute scale A is \( \frac{273.16}{200} \)
or \( 1A = \frac{273.16}{200}K \)
And the size of one degree of Kelvin scale on absolute scale B is \( \frac{273.16}{350} \)
or \( 1B = \frac{273.16}{350}K \)
Therefore, the value of temperature \( T_{A} \) on the absolute scale is \( \frac{273.16}{200} \times T_{A} \)
And the value of temperature \( T_{B} \) on the absolute scale is \( \frac{273.16}{350} \times T_{B} \)
Since \( T_{A} \) and \( T_{B} \) are at the same temperature.
\( \implies \frac{273.16}{200} \times T_{A} = \frac{273.16}{350} \times T_{B} \)
or \( T_{A} = \frac{200}{350} \times T_{B} \)
or \( T_{A} = \frac{200}{350}T_{B} = \frac{4}{7}T_{B} \).
In simple words: The ratio of the reference points on two absolute temperature scales tells you how one degree on one scale compares to a degree on the other. This ratio then helps you link any temperature on one scale to the equivalent temperature on the other.

Exam Tip: When dealing with new or arbitrary scales, relate them back to a known standard (like Kelvin's triple point) to establish conversion factors between them.

 

Question 3. according to the approximate law: \( R = R_{0}[1 + 5 \times 10^{-3}(T - T_{0})] \) The resistance is 101.6 Ω at the triple-point of water, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?
Answer: Here, \( R_{1} = 101.6 \Omega \) at \( T_{1} = 273.16 K \)
\( R_{2} = 165.5 \Omega \) at \( T_{2} = 600.5 K \)
Let \( R_{0} \) be the resistance at \( T_{0} \).
And let \( R_{3} = 123.4 \Omega \) at Temp. \( T_{3} = ? \)
The given law for electrical resistance as a function of temperature is stated as:
\( R = R_{0}[1 + 5 \times 10^{-3}(T - T_{0})] \)

(i) \( 101.6 = R_{0}[1 + 5 \times 10^{-3}(273.16 - T_{0})] \)

(ii) \( 165.5 = R_{0}[1 + 5 \times 10^{-3}(600.5 - T_{0})] \)

(iii) \( \frac{165.5}{101.6} = \frac{1+5 \times 10^{-3}\left(600.5-T_{0}\right)}{1+5 \times 10^{-3}\left(273.16-T_{0}\right)} \)
or \( 1 + 5 \times 10^{-3}(600.5 - T_{0}) = 1.629[1 + 5 \times 10^{-3}(273.16 - T_{0})] \)
or \( 1.629(1 + 1.366 - 0.005 T_{0}) = 1.629[1 + 5 \times 10^{-3}(273.16 - T_{0})] \)
or \( 3.854 - 0.008 T_{0} = 4.003 - 0.005 T_{0} \)
or \( 0.003T_{0} = 0.149 \)
or \( T_{0} = -49.67 K \)
Thus, from (ii), \( R_{0} = \frac{101.6}{1+0.005(273.16+49.67)} \)
\( = \frac{101.6}{2.614} = 38.87 \Omega \)
So, \( 123.4 = 38.87 \{1 + 0.005 [T - (-49.67)]\} \)
or \( T + 49.67 = (\frac{123.4}{38.87} - 1) \frac{1}{0.005} \)
or \( T = 434.94 - 49.67 = 385 K \).
Aliter:
\( R = R_{0}[1 + 5 \times 10^{-3}(T - T_{0})] \)
Also, we recognize that \( R_{1} = R_{0} [1 + \alpha(T - T_{0})] \)
or \( T-T_{0} = \frac{R-R_{0}}{\alpha R_{0}} \)

\( \implies T_{1} - T_{0} = \frac{R_{1}-R_{0}}{\alpha R_{0}} \) ..... (1)

\( T_{2}-T_{0} = \frac{R_{2}-R_{0}}{\alpha R_{0}} \) ..... (2)

\( \implies \frac{T_{2}-T_{0}}{T_{1}-T_{0}} = \frac{R_{2}-R_{0}}{R_{1}-R_{0}} \) ..... (3)
Here \( R_{0} \) is the resistance at \( T_{0} \).
Specifically, \( T_{0} = 273.16 K \), \( R_{0} = 101.16\Omega \), \( T_{1} = 600.5 K \), \( R_{1} = 165.5 \Omega \), \( T_{2} = ? \), \( R_{2} = 123.4 \Omega \).
..... (4)
Thus, from (3) and (4), we obtain
\( T_{2}-T_{0} = (\frac{123.4-101.6}{165.5-101.6}) (600.5 - 273.16) \)
\( = 384.88 = 385 K \).
In simple words: This problem asks us to find the temperature based on a material's electrical resistance. We use a formula that connects resistance to temperature, then substitute the known resistance values at specific temperatures to find unknown values, sometimes using an alternative method for verification.

Exam Tip: For problems involving resistance-temperature relationships, ensure you correctly identify the known and unknown variables and consistently apply the given formula or a derived ratio for calculations.

 

Question 4. Answer the following:
(a) The triple point Of water is a standard fixed point in modern thermometer. Why? What is wrong in taking melting point of ice and the boiling point of water as standard fixed points (as was originally done in Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0° C and 100° C respectively. On the absolute scale one of the fixed points is the triple point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale.
(c) The absolute temperature (Kelvin scale) T is related to the temperature T on the Celsius scale by \( t_{c} = T - 273.15 \). Why do we have 273.15 in this relation and not 273.16?
(d) What is the temperature of triple point of water on an absolute scale whose unit interval size is equal that of the Fahrenheit scale?
Answer:
(a) The triple point of water is a unique state because it happens at only one particular combination of pressure and temperature. This makes it easily replicable, which is why it is used as a standard reference point in modern thermometry. In contrast, the melting point of ice and the boiling point of water change with pressure. They are also very sensitive to dissolved impurities, which is why they were replaced as fixed points.
(b) The other fixed point on the absolute temperature scale, besides the triple point of water, is absolute zero (0 K). This is the theoretical temperature where the volume and pressure of an ideal gas would become zero.
(c) The triple point of water is \( 0.01^{\circ}C \), not \( 0^{\circ}C \). To ensure that a Celsius degree matches a Kelvin degree, 273.15 is assigned instead of 273.16. The ice point on the Kelvin scale is 273.15 K, and the corresponding ice point on the Celsius scale is \( 0^{\circ}C \). If we used 273.16, the ice point on the Celsius scale would be \( -0.01^{\circ}C \), which is incorrect.
(d) The Fahrenheit scale has 212 - 32 = 180 divisions. We also know that the absolute scale has 100 unit interval divisions. Therefore, the triple point of water on an absolute scale with 180 divisions is given by:
\( T = \frac{273.16}{100} \times 180 = 491.69 \).
In simple words: The triple point of water is a reliable temperature reference because it only occurs under very specific conditions, unlike boiling or melting points which can change. Absolute zero is the lowest possible temperature. The 273.15 value in Kelvin-Celsius conversion ensures consistency with the triple point. If a new scale has 180 divisions like Fahrenheit, then water's triple point would be 491.69 on that new scale.

Exam Tip: Understand the fundamental definitions of fixed points in thermometry. The key advantage of the triple point of water is its unique and reproducible nature, making it a highly stable reference.

 

Question 5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason for slightly different answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Answer:
(a) Let T be the melting point of sulphur. We also know that the triple point of water \( T_{tr} = 273.16 K \).
For thermometer A, \( P_{tr} = 1.250 \times 10^{5} Pa \), \( P = 1.797 \times 10^{5} Pa \), \( T = ? \)
Using the relation,
\[ \frac{T}{T_{tr}} = \frac{P}{P_{tr}} \]
\[ T_{A} = \frac{P}{P_{tr}} \times T_{tr} \]
\[ T_{A} = \frac{1.797 \times 10^{5}}{1.250 \times 10^{5}} \times 273.16 \]
\( = 392.69 K \).
For thermometer B, \( P_{tr} = 0.200 \times 10^{5} Pa \), \( P = 0.287 \times 10^{5} Pa \)
\[ T_{B} = T_{tr} \times \frac{P}{P_{tr}} \]
\[ = 273.16 \times \frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \]
or \( T_{B} = 391.98 K \).
(b) The slight difference in the melting point of sulphur measured by the two thermometers is because, in actual practice, gases do not behave exactly as ideal gases. To reduce this difference, measurements should be taken at lower and lower pressures. Then, a plot between the measured temperature and the absolute pressure of the gas at the triple point should be extrapolated to determine the temperature as the pressure approaches zero (i.e., \( P \rightarrow 0 \)). In this limit, gases behave more like ideal gases.
In simple words: Thermometers A and B give slightly different readings for sulphur's melting point because real gases don't perfectly follow ideal gas rules. To get more accurate results, you need to take readings at very low pressures and then predict what the temperature would be if the pressure was zero, where gases act most ideally.

Exam Tip: Remember that ideal gas thermometers rely on the assumption of ideal gas behavior. Real gases deviate, especially at higher pressures, leading to small discrepancies. Extrapolating to zero pressure minimizes these deviations.

 

Question 6. A steel tape 1 m long is correctly calibrated for a temperature of 27.0°C. The length of steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is. 45.0°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when temperature is 27.0°C? Coefficient of linear expansion of steel = \( 1.20 \times 10^{-5}K^{-1} \)?
Answer: A steel tape is calibrated at \( 27^{\circ}C \). This means it accurately measures length at \( 27^{\circ}C \), so a 1 cm mark on the tape has the correct dimension.
The length of the steel rod at \( 27^{\circ}C \) is \( L_{0} = 63 \text{ cm} \), and the coefficient of linear expansion \( \alpha = 1.2 \times 10^{-5}K^{-1} \).
If \( \Delta L \) is the increase in the length of the steel rod or steel tape when the temperature rises from \( 27^{\circ}C \) to \( 45^{\circ}C \),
Then \( \Delta L = \alpha L_{0}\Delta T \)
\( = 1.2 \times 10^{-5} \times 63 \times (45 - 27) \text{ cm} \)
\( = 1.2 \times 10^{-5} \times 63 \times 18 \text{ cm} \)
\( = 0.0136 \text{ cm} \).
Thus, the actual length of the steel rod at \( 45^{\circ}C \)
\( = L_{0} + \Delta L \)
\( = 63 + 0.0136 \)
\( = 63.0136 \text{ cm} \)
The data is accurate only up to three significant figures. The actual length of the rod is 63.0 cm, but the change in length is 0.0136 cm. On a day when the temperature is \( 27^{\circ}C \), the size of the 1 cm mark on the steel tape will be exactly 1 cm (because of the calibration of the steel tape at \( 27^{\circ}C \)).
Thus, the length of the rod at \( 27^{\circ}C = 63.0 \times 1 = 63.0 \text{ cm} \).
Aliter:
The length of the steel tape at \( 27^{\circ}C \) is 100 cm.
So, \( L_{0} = 100 \text{ cm} \)
\( T = 27^{\circ}C \).
Thus, \( L' = \) length at \( 45^{\circ}C = L_{0} + \alpha L_{0} \Delta T \)
\( = 100 + 1.2 \times 10^{-5} \times 100 \times (45 - 27) \)
\( = 100 + 1.2 \times 10^{-5} \times 100 \times 18 \)
\( = 100 + 0.0216 \)
\( = 100.0216 \text{ cm} \)
Therefore, the length of 63 cm measured by this tape at \( 45^{\circ}C \) will actually be
\( = \frac{100.0216}{100} \times 63 = 63.0136 \text{ cm} \).
In simple words: First, calculate how much the tape expands at the higher temperature. Then, add this expansion to the measured length to find the rod's true length on the hot day. The rod's length at its calibration temperature is simply the measured length.

Exam Tip: Always remember to account for the expansion of the measuring tool itself when the temperature differs from its calibration temperature. The formula \( \Delta L = \alpha L_{0}\Delta T \) is fundamental here.

 

Question 7. A large steel wheel is to be fitted on a shaft of the same material. At 27°C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range?
Answer: The coefficient of linear expansion of steel is \( \alpha_{\text{steel}} = 1.20 \times 10^{-5}K^{-1} \)
Initial diameter of the shaft, \( L_{1} = 8.70 \text{ cm} \)
Diameter of the hole, \( L_{2} = 8.69 \text{ cm} \)
Initial temperature, \( T_{1} = 27^{\circ}C = 273 + 27 = 300 K \)
We need to find the final temperature, \( T_{2} = ? \)
When the shaft cools to temperature \( T_{2} \) such that its diameter becomes \( L_{2} = 8.69 \text{ cm} \), the wheel can easily slip on the shaft. Using the linear expansion formula:
\( L_{2} = L_{1}[1 + \alpha(T_{2} - T_{1})] \)
\( 8.69 = 8.70[1 + 1.20 \times 10^{-5}(T_{2} - 300)] \)
\( \frac{8.69}{8.70} = 1 + 1.20 \times 10^{-5}(T_{2} - 300) \)
\( 0.99885 = 1 + 1.20 \times 10^{-5}(T_{2} - 300) \)
\( -0.00115 = 1.20 \times 10^{-5}(T_{2} - 300) \)
\( T_{2} - 300 = \frac{-0.00115}{1.20 \times 10^{-5}} \)
\( T_{2} - 300 = -95.83 \)
\( T_{2} = 300 - 95.83 = 204.17 K \)
Converting to Celsius:
\( T_{2} = 204.17 - 273.15 = -68.98^{\circ}C \)
or \( T_{2} \approx -69^{\circ}C \).
In simple words: To make the shaft fit the wheel, we need to cool the shaft until its diameter shrinks to match the wheel's hole. We use the material's expansion coefficient and the initial dimensions to calculate the exact temperature required for the shaft to shrink enough.

Exam Tip: Remember that cooling causes contraction, and the linear expansion formula also applies to changes in diameter. Pay close attention to unit consistency (K vs. °C) during calculations.

 

Question 8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = \( 1.70 \times 10^{-5}{}^{\circ}C^{-1} \)?
Answer: Here, the initial temperature is \( t_{1} = 27^{\circ}C \)
The final temperature is \( t_{2} = 227^{\circ}C \)
The coefficient of linear expansion \( \alpha \) for copper is \( 1.70 \times 10^{-5}{}^{\circ}C^{-1} \)
Therefore, the change in temperature \( \Delta t = t_{2} - t_{1} = 227 - 27 = 200.0^{\circ}C \)
The diameter of the hole at \( 27^{\circ}C \) is \( d_{1} = 4.24 \text{ cm} \).
Let the diameter of the hole at \( 227^{\circ}C \) be \( d_{2} \).
We need to find the change in diameter, \( \Delta d = d_{2} - d_{1} = ? \)
If \( \beta \) is the coefficient of superficial (area) expansion of copper,
Then \( \beta = 2\alpha = 2 \times 1.7 \times 10^{-5} = 3.4 \times 10^{-5}{}^{\circ}C^{-1} \)
The initial area of the hole is \( S_{1} = \frac{\pi d_{1}^{2}}{4} = \frac{\pi}{4} \times (4.24)^{2} = 4.494\pi \text{cm}^{2} \)
The final area of the hole is \( S_{2} = S_{1}(1 + \beta \Delta t) \)
\( = 4.494 \pi (1 + 3.40 \times 10^{-5} \times 200) \)
\( = 4.494 \pi (1 + 0.0068) \)
\( = 4.494 \pi \times 1.0068 = 4.5246 \pi \text{cm}^{2} \)
So, \( \frac{\pi d_{2}^{2}}{4} = 4.5246 \pi \)
\( d_{2}^{2} = 4 \times 4.5246 = 18.0984 \)
\( d_{2} = \sqrt{18.0984} = 4.2542 \text{ cm} \)
Therefore, \( \Delta d = d_{2} - d_{1} = 4.2542 - 4.24 \)
\( = 0.0142 \text{ cm} \)
or \( \Delta d = 1.42 \times 10^{-2} \text{ cm} \).
In simple words: When a copper sheet with a hole gets warmer, the hole itself also expands. We use the copper's linear expansion rate and the temperature change to figure out how much the hole's diameter will grow.

Exam Tip: Remember that holes in materials expand and contract just like the material itself. The coefficient of linear expansion applies to dimensions like diameter, and area expansion (beta = 2*alpha) can be used for changes in area.

 

Question 9. A brass wire 1.8 m long at 27°C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of -39°C, what is the tension developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of brass = \( 2.0 \times 10^{-5}K^{-1} \). Young's modulus of brass = \( 0.91 \times 10^{11} Pa \)?
Answer: Here, the initial length \( L_{1} = 1.8 \text{ m} \), initial temperature \( t_{1} = 27^{\circ}C \), final temperature \( t_{2} = -39^{\circ}C \).
Therefore, the change in temperature \( \Delta t = t_{2} - t_{1} = -39 - 27 = -66^{\circ}C \).
For brass, the coefficient of linear expansion is \( \alpha = 2 \times 10^{-5}K^{-1} \).
Young's modulus is \( Y = 0.91 \times 10^{11} Pa \).
The diameter of the wire \( d = 2.0 \text{ mm} = 2.0 \times 10^{-3} \text{ m} \).
If \( A \) is the area of the cross-section of the wire, then
\( A = \frac{\pi d^{2}}{4} = \frac{\pi}{4} \times (2.0 \times 10^{-3})^{2} \)
\( = 3.142 \times 10^{-6}m^{2} \).
If \( F \) is the tension developed in the wire, then using the relation for Young's modulus:
\( Y = \frac{F/A}{\Delta L/L_{1}} \)
We can rearrange this to find \( F \):
\( F = \frac{Y A \Delta L}{L_{1}} \)
First, calculate the change in length \( \Delta L \) due to cooling:
\( \Delta L = L_{1}\alpha\Delta t = 1.8 \times 2 \times 10^{-5} \times (-66) \)
\( = -0.002376 \text{ m} \approx -0.0024 \text{ m} \)
The negative sign shows that the length decreases.
Now, putting the values of \( Y \), \( A \), \( \Delta L \) (magnitude), and \( L_{1} \) into the equation for \( F \), we get
\( F = \frac{0.91 \times 10^{11} \times 3.142 \times 10^{-6} \times 0.0024}{1.8} \)
\( F \approx 381.18 N \)
So, \( F \approx 381 N \) or \( F \approx 3.81 \times 10^{2} N \).
In simple words: When a taut brass wire cools down, it tries to shrink. But because it's held by rigid supports, it can't shrink, so this creates a pulling force, or tension. We calculate this tension using the wire's length, how much it would normally shrink, its stiffness (Young's modulus), and its cross-sectional area.

Exam Tip: For problems involving thermal stress or tension, remember that Young's modulus relates stress to strain. First calculate the thermal strain (\( \alpha\Delta T \)), and then use the formula \( F = Y A (\alpha\Delta T) \) to find the force (tension).

 

Question 10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40°C? Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (coefficient of linear expansion of brass = \( 2.0 \times 10^{-5}K^{-1} \), steel = \( 1.2 \times 10^{-5}K^{-1} \).)
Answer: For the brass rod:
The coefficient of linear expansion \( \alpha = 2.0 \times 10^{-5}K^{-1} \)
Initial length \( L_{1} = 50 \text{ cm} \)
Initial temperature \( t_{1} = 40^{\circ}C \)
Final temperature \( t_{2} = 250^{\circ}C \).
Thus, the change in temperature \( \Delta t = t_{2} - t_{1} = 250 - 40 = 210^{\circ}C \).
If \( L_{2} \) is its length at \( 250^{\circ}C \), then
\( L_{2} = L_{1}(1 + \alpha\Delta t) \)
\( = 50(1 + 2 \times 10^{-5} \times 210) \)
\( = 50(1 + 0.0042) \)
\( = 50.21 \text{ cm} \)
Therefore, the change in length for brass \( \Delta L_{\text{brass}} = L_{2} - L_{1} = 50.21 - 50 = 0.21 \text{ cm} \).
For the steel rod:
Initial temperature \( t'_{1} = 40^{\circ}C \)
Final temperature \( t'_{2} = 250^{\circ}C \)
Coefficient of linear expansion \( \alpha = 1.2 \times 10^{-5}K^{-1} \)
Initial length \( L'_{1} = 50.0 \text{ cm} \)
Change in temperature \( \Delta t' = t'_{2} - t'_{1} = 250 - 40 = 210^{\circ}C \).
If \( L'_{2} \) is the length of the steel rod at \( 250^{\circ}C \), then
\( L'_{2} = L'_{1}(1 + \alpha \Delta t') \)
\( = 50(1 + 1.2 \times 10^{-5} \times 210) \)
\( = 50(1 + 0.00252) \)
\( = 50.126 \text{ cm} \).
Therefore, the change in length for steel \( \Delta L_{\text{steel}} = L'_{2} - L'_{1} = 50.126 - 50 = 0.126 \text{ cm} \).
This can be rounded to \( 0.13 \text{ cm} \).
Thus, the total length of the combined rod at \( 250^{\circ}C = L_{2} + L'_{2} \)
\( = 50.21 + 50.126 \)
\( = 100.336 \text{ cm} \)
And the initial length of the combined rod at \( 40^{\circ}C = L_{1} + L'_{1} \)
\( = 50 + 50 \)
\( = 100 \text{ cm} \)
Therefore, the total change in length of the combined rod \( = 100.336 - 100 \)
\( = 0.336 \text{ cm} \)
This can be rounded to \( 0.34 \text{ cm} \).
No 'thermal stress' is created at the junction because the rods are allowed to expand freely.
In simple words: First, calculate how much each rod (brass and steel) stretches when heated from 40°C to 250°C. Then, add these individual length changes to find the total change in the combined rod's length. Since the rods can expand freely, there's no stress where they meet.

Exam Tip: When different materials are joined and heated, calculate the expansion for each material separately. If the ends are free to expand, no thermal stress develops; stress only occurs if expansion is constrained.

 

Question 11. The coefficient of volume expansion of glycerine is \( 49 \times 10^{-5}K^{-1} \). What is the fractional change in its density for a 30°C rise in temperature?
Answer: Here, the coefficient of volume expansion \( \gamma = 49 \times 10^{-5}K^{-1} \).
The rise in temperature, \( \Delta t = 30^{\circ}C \).
Let \( V_{0} \) be the initial volume of glycerine at \( 0^{\circ}C \).
If \( V_{t} \) is its volume after the temperature rise,
Then \( V_{t} = V_{0}(1 + \gamma\Delta t) \)
\( = V_{0} (1 + 49 \times 10^{-5} \times 30) \)
\( = V_{0}(1 + 0.01470) = 1.01470V_{0} \)
or \( \frac{V_{0}}{V_{t}} = \frac{1}{1.01470} \) ..... (i)
Let \( \rho_{0} \) and \( \rho_{t} \) be the initial and final densities of the glycerine.
Then the initial density, \( \rho_{0} = \frac{m}{V_{0}} \)
And the final density, \( \rho_{t} = \frac{m}{V_{t}} \), where \( m \) is the mass of glycerine.
The fractional change in density is \( \frac{\Delta\rho}{\rho_{0}} = \frac{\rho_{t} - \rho_{0}}{\rho_{0}} \)
\[ \frac{\Delta\rho}{\rho_{0}} = \frac{\rho_{t} - \rho_{0}}{\rho_{0}} = \frac{m/V_{t} - m/V_{0}}{m/V_{0}} = \frac{1/V_{t} - 1/V_{0}}{1/V_{0}} = \frac{V_{0}}{V_{t}} - 1 \]
\[ \frac{\Delta\rho}{\rho_{0}} = (\frac{1}{1.01470} - 1) = -0.0145 \]
The negative sign shows that the density decreases as the temperature rises.
Therefore, \( \frac{\Delta\rho}{\rho_{0}} = 0.0145 = 1.45 \times 10^{-2} \)
This can be rounded to \( 1.5 \times 10^{-2} \).
In simple words: When glycerine gets warmer, its volume increases, which means its density decreases. We calculate this density change by using its volume expansion coefficient and the temperature rise. The answer tells us by what fraction the density has fallen.

Exam Tip: Remember that density is inversely proportional to volume for a given mass. Volume expansion leads to a decrease in density, and the fractional change in density is approximately equal to minus the coefficient of volume expansion times the temperature change (\( -\gamma\Delta t \)).

 

Question 12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of the power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = \( 0.91 \text{ Jg}^{-1}K^{-1} \)?
Answer: Here, the power of the machine is \( P = 10\text{ kW} = 10 \times 10^{3}W \).
The time for which the machine is used is \( t = 2.5 \text{ minutes} = 2.5 \times 60\text{s} = 150\text{s} \).
The mass of the aluminium block is \( M = 8 \text{ kg} \).
The specific heat of aluminium is \( C = 0.91 \text{ Jg}^{-1}K^{-1} \).
If \( Q \) is the total energy used by the drilling machine, then
\( Q = P \times t \)
\( = 10 \times 10^{3} \times 150 \)
\( = 15 \times 10^{5}J \)
It is stated that 50% of the energy is lost to the surroundings or used in heating the machine itself.
Therefore, if \( Q' \) is the energy transferred to the aluminium block, then
\( Q' = 50\% \text{ of } Q = \frac{50}{100} \times 15 \times 10^{5} \)
\( = 7.5 \times 10^{5} J \).
We know that the heat absorbed by the block is given by \( Q' = m C \Delta t \), where \( \Delta t \) is the rise in temperature of the block.
Rearranging to find \( \Delta t \):
\( \Delta t = \frac{Q'}{mC} \)
\( = \frac{7.5 \times 10^{5}}{8 \times 0.91 \times 10^{3}} \)
\( = \frac{750000}{7280} \)
\( \approx 103.02^{\circ}C \)
So, the rise in temperature is approximately \( 103^{\circ}C \).
In simple words: We first calculate the total energy the drilling machine uses. Since only half of this energy heats the aluminium block, we take 50% of the total. Then, using the block's mass and specific heat, we figure out how much its temperature will go up from this absorbed energy.

Exam Tip: When solving problems involving energy transfer, always check for efficiency or energy loss. Only the energy effectively transferred to the object contributes to its temperature change, calculated using \( Q = mc\Delta T \).

 

Question 13. A copper block of mass 2.5 kg is heated in furnace to a temperature of 500°C and then placed on a large ice block. What in the maximum amount of ice that can melt? (Specific heat of copper = \( 0.39 \text{ Jg}^{-1}K^{-1} \); heat of fusion of water = \( 335 \text{ Jg}^{-1} \))
Answer: Here, the mass of the copper block is \( m_{1} = 2.5 \text{ kg} \).
The specific heat of copper is \( C = 0.39 \text{ Jg}^{-1}K^{-1} = 0.39 \times 10^{3} \text{ J kg}^{-1}K^{-1} \).
The temperature of the furnace, \( \Delta \Theta = 500^{\circ}C \).
The latent heat of fusion for water is \( L = 335 \text{ Jg}^{-1} = 335 \times 10^{3} \text{ J Kg}^{-1} \).
If \( Q \) is the heat absorbed by the copper block, then

(i) \( Q = m_{1}C\Delta \Theta \)
Let \( m_{2}(\text{kg}) \) be the mass of ice melted when the copper block is placed on it.

(ii) \( Q = m_{2}L \)
From (i) and (ii), we equate the heat values:
\( m_{1}C\Delta \Theta = m_{2}L \)
or \( m_{2} = \frac{m_{1} C \Delta \Theta}{L} \)
\( = \frac{2.5 \times 0.39 \times 10^{3} \times 500}{335 \times 10^{3}} \)
\( = \frac{2.5 \times 0.39 \times 500}{335} \)
\( = \frac{487.5}{335} \)
\( = 1.455 \text{ kg} \)
So, the maximum amount of ice that can melt is approximately \( 1.5 \text{ kg} \).
In simple words: We calculate the total heat lost by the hot copper block. This heat is then used to melt the ice. By dividing the heat lost by the copper block by the latent heat of fusion for ice, we find out the maximum amount of ice that can be melted.

Exam Tip: For calorimetry problems, the heat lost by the hotter object equals the heat gained by the colder object. When phase changes occur (like melting ice), remember to use the latent heat of fusion in addition to specific heat capacity for temperature changes.

 

Question 14. In an experiment on the specific heat of a metal a 0.20 kg block of the metal at 150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm³ of at 27°C. The final temperature is 40°C. Compute the specific heat of the metal. If beat losses to the surrounding are not negligible is your answer greater or smaller than the actual value for specific heat of the metal?
Answer: Here, the mass of the metal block is \( m_{1} = 0.20 \text{ kg} \).
The initial temperature of the metal block is \( \theta_{1} = 150^{\circ}C \).
The initial temperature of the calorimeter and water is \( \theta_{2} = 27^{\circ}C \).
The final temperature of the mixture is \( \theta = 40^{\circ}C \).
The mass of water in the calorimeter, \( m_{2} = V\rho = 150 \times 10^{-6} \times 10^{3}\text{kg} \)
\( = 0.150 \text{ kg} \).
The water equivalent of the calorimeter is \( W = 0.025 \text{ kg} = 25 \times 10^{-3} \text{ kg} \).
Let \( C_{1} (\text{J kg}^{-1} \text{ K}^{-1}) \) be the specific heat of the metal block.
Therefore, the fall in temperature of the metal block, \( \Delta\theta = \theta_{1} - \theta = 150 - 40 = 110^{\circ}C \).
The increase in the temperature of the water and calorimeter, \( \Delta\Theta = \theta - \theta_{2} = 40 - 27 = 13^{\circ}C \).
If \( Q_{1} \) is the heat lost by the metal block, then
\( Q_{1} = m_{1}C_{1}\Delta\theta \)
\( = 0.20 \times C_{1} \times 110 = 22C_{1} \)
Also, let \( C_{2} \) be the specific heat of water, \( C_{2} = 4.2 \times 10^{3} \text{ J kg}^{-1} \text{ K}^{-1} \).
If \( Q_{2} \) is the heat gained by the water and the calorimeter, then
\( Q_{2} = (m_{2} + W)C_{2}\Delta\Theta \)
\( = (0.150 + 0.025)C_{2} \times 13 \)
\( = (175 \times 10^{-3})C_{2} \times 13 \)
\( = 175 \times 10^{-3} \times 4.2 \times 10^{3} \times 13 \)
\( = 9555 J \).
According to the principle of Calorimetry, heat lost equals heat gained:
\( Q_{1} = Q_{2} \)
\( 22C_{1} = 9555 \)
\( C_{1} = \frac{9555}{22} \)
\( C_{1} = 434.32 \text{ J kg}^{-1} \text{ K}^{-1} \)
This can also be expressed as \( \frac{434.32}{1000} \text{ J kg}^{-1} \text{ K}^{-1} = 0.434 \text{ J g}^{-1} \text{ K}^{-1} \).
If heat is lost to the surroundings, the calculated \( C_{1} \) value would be smaller than the actual value, because less heat would appear to be gained by the water/calorimeter.
In simple words: We find the heat lost by the hot metal block and the heat gained by the water and calorimeter as they reach a final temperature. By equating these, we can calculate the metal's specific heat. If some heat escaped to the surroundings, our calculated specific heat for the metal would appear lower than its true value.

Exam Tip: In calorimetry experiments, always consider potential heat losses to the surroundings. If ignored, the calculated specific heat of the object being tested will be inaccurately low because some of its lost heat wasn't recorded as gained by the calorimeter/water.

 

Question 15. Given below are observations on molar specific heats at room temperatures of some common gases?

The measured molar specific heats of these gases are markedly different from thosefor monoatomic gases (typically molar specific heat of a monoatomic gas is \( 2.92 \text{ cal/mol K} \)). Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?
Answer: Since all the specified gases are diatomic, they possess other degrees of freedom besides translational motion (such as rotational and vibrational modes). To raise the temperature of these gases by a particular amount, heat must be supplied to increase the average energy in all these modes. Consequently, the molar specific heat of diatomic gases is greater than that of monoatomic gases.
It can be shown that if only rotational modes of motion are considered, the molar specific heat of diatomic gases is approximately \( \frac{5}{2} R = 2.5 \times 1.985 = 4.962 \text{ Cal mol}^{-1} \text{ K}^{-1} \). This value aligns well with the observations for all the gases listed in the table, except for chlorine. The higher value of molar specific heat for chlorine suggests that, in addition to translational and rotational modes, vibrational modes are also active in chlorine at room temperature.
In simple words: Diatomic gases need more heat to warm up than single-atom gases because their energy can go into rotating and vibrating, not just moving around. Chlorine has an even higher specific heat, which means its atoms are also vibrating a lot at room temperature, using up more energy.

Exam Tip: The molar specific heat of a gas depends on its degrees of freedom (translational, rotational, vibrational). Diatomic gases generally have higher specific heats than monoatomic ones due to rotational modes; exceptionally high values (like chlorine) indicate active vibrational modes even at room temperature.

 

Question 16. Answer the following questions based on the P – T phase diagram of CO2 as shown in fig. here?

(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?
(c) What are the critical temperature and pressure for CO2? What is their significance?
(d) Is CO2 solid, liquid or gas at:

• - 70°C under 1 atm,


• - 60°C under 10 atm,


• 15°C under 56 atm?

Answer:
(a) The solid, liquid, and vapour phases of \( CO_{2} \) can all exist together in equilibrium at its triple point O. From the diagram, this point corresponds to a temperature of \( -56.6^{\circ}C \) and a pressure of \( 5.11 \text{ atm} \).
(b) Looking at the vaporization curve (I) and the fusion curve (II), we observe that both the boiling point and the fusion point of \( CO_{2} \) decrease as the pressure decreases. This means lower pressure leads to lower phase transition temperatures.
(c) For \( CO_{2} \), the critical pressure \( P_{c} = 73.0 \text{ atm} \) and the critical temperature \( T_{c} = 31.1^{\circ}C \). The significance of these values is that above the critical temperature, \( CO_{2} \) cannot be liquefied, no matter how much pressure is applied to it.
(d)

• At \( -70^{\circ}C \) under 1 atm, \( CO_{2} \) is in the vapour phase because this point lies within the vapour region of the phase diagram.


• At \( -60^{\circ}C \) under 10 atm, \( CO_{2} \) is in the solid phase because this point lies within the solid region of the phase diagram.


• At \( 15^{\circ}C \) under 56 atm, \( CO_{2} \) is in the liquid phase because this point falls within the liquid region of the phase diagram.

In simple words: The triple point shows where all three states (solid, liquid, gas) of \( CO_{2} \) exist together. Lowering the pressure reduces the boiling and melting points. The critical point is the temperature and pressure above which \( CO_{2} \) cannot become a liquid. By checking the diagram, you can tell if \( CO_{2} \) is solid, liquid, or gas at different temperatures and pressures.

Exam Tip: When analyzing phase diagrams, locate the triple point and critical point first, as these are key reference points. Understand how moving across phase boundary lines corresponds to phase transitions, and remember that critical temperature defines the upper limit for liquefaction.

 

Question 17. Answer the following questions based on the P – T phase diagram of CO2 as shown in fig. of Q. 11.16.
(a) CO2 at 1 atm pressure and temperature – 60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature – 65°C as it is heated up to room temperature at constant pressure.
(d) CO2 is heated to a temperature 70°C and compressed isothermally. What changes in its properties do you expect to observe?
Answer:
(a) No, CO2 at 1 atm pressure and – 60°C is vapor. If it is compressed isothermally, meaning when pressure increases without changing the temperature, it will go to the solid phase directly without passing through the liquid phase.
(b) CO2 at 4 atm pressure and at room temperature (let's say 25°C) is vapor. If it cools at constant pressure, it will again condense directly to solid without going through the liquid phase. The horizontal line passing through the initial point crosses only the sublimation curve (III).
(c) CO2 at 10 atm pressure and at – 60°C is solid. As CO2 is heated at constant pressure, it will turn into a liquid state and then into the vapor phase. This happens because the horizontal line from the starting point crosses both the fusion and vaporization curves. The fusion and boiling points can be found from the points where the horizontal line in the P – T diagram intersects the respective curves.
(d) It will not show any clear phase change to the liquid phase. However, CO2 gas will gradually move further from ideal gas behavior as its pressure increases.
In simple words: When CO2 is compressed or cooled under certain conditions, it can change directly between solid and gas without becoming a liquid. The phase diagram helps us see these changes.

Exam Tip: For phase diagrams, accurately identify the regions (solid, liquid, gas) and the phase boundary lines (fusion, vaporization, sublimation) to correctly interpret phase changes under different conditions.

 

Question 18. A child running a temperature of 101°F is given an antipyrin (i.e. medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is thought down to 98°F in 20 min., what is the average rate of extra evaporation caused by the drug. Assume the evaporation mechanism to be the only way by which heat is lost. The mass of child is 30 kg. The specific heat of the human body is approximately the same as that of water and latent heat of evaporation of water at that temperature is about 580 cal g-1.
Answer:
Here, mass of the child, \( m = 30 \) kg
Fall in temperature, \( \Delta T = T_1 - T_2 = 101^\circ F - 98^\circ F = 3^\circ F \)
\( = 3 \times \frac{5}{9}^\circ C \)
or \( \Delta T = \frac{5}{3}^\circ C \)
Specific heat of human body (sp. heat of water) \( C = 4.2 \times 10^3 \) J kg\(^{-1}\)C\(^{-1}\)
Latent heat of vaporization \( = 580 \) cal g\(^{-1}\)
\( = 580 \times 4.2 \times 10^3 \) J kg\(^{-1}\)C\(^{-1}\)
Let \( m' \) gram of sweat evaporates from the body of the child in 20 min, i.e. \( t = \) time taken by sweat to evaporate \( = 20 \) min.
If Q be the heat needed, then
\( Q = m'L \)
\( = m \times 580 \times 4.2 \times 10^3 \) J .................... (1)
Also let Q be the heat lost by the child in the form of evaporation of sweat, then
\( Q = mC\Delta T \)
\( = 30 \times 4.2 \times 10^3 \times 5 \) J
\( = 2.10 \times 10^5 \) J .................... (2)
From (1) and (2), we get
\( m \times 580 \times 4.2 \times 10^3 = 2.1 \times 10^5 \)
\( m' = \frac{2.1 \times 10^5}{580 \times 4.2 \times 10^3} \)
\( = \frac{2.1 \times 10^5}{2436 \times 10^3} \)
\( = \frac{210}{24360} \)
\( = 0.00862 \) kg
Rate of evaporation of sweat \( = \frac{m'}{t} = \frac{0.00862}{20} \)
\( = 0.000431 \) kg min\(^{-1}\)
\( = 0.431 \) g min\(^{-1}\).
In simple words: We calculated how much heat the child loses as their fever drops. Then, using the latent heat of sweat, we figured out how much sweat needed to evaporate to remove that heat over 20 minutes. Finally, we divided the amount of sweat by the time to find the average evaporation rate.

Exam Tip: Remember to convert all units to be consistent (e.g., Fahrenheit to Celsius, minutes to seconds, calories to joules) before beginning calculations to prevent errors. Ensure you use the correct latent heat value for evaporation.

 

Question 19. A 'thermacole' icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6h. The outside temperature is 45°C and coefficient of thermal conductivity of thermacole is 0.01 J s-1m-10K-1. |Heat of fusion of water = 335 × 103J kg-1
Answer:
Here,
Area of 6 faces of cube \( = 6 \times 30 \times 30 \) cm\(^2\)
\( = 6 \times 900 \times 10^{-4} \) m\(^2\)
\( = 0.54 \) m\(^2\)
Distance, \( d = 5.0 \) cm \( = 5.0 \times 10^{-2} \) m
Total mass of ice, \( M = 4 \) kg
Time, \( t = 6 \)h \( = 6 \times 60 \times 60 \) s \( = 21600 \) s
\( \theta_1 = \) temperature outside the box \( = 45^\circ \)C
\( \theta_2 = \) temperature inside the box \( = 0^\circ \)C
\( \Delta\theta = \theta_1 - \theta_2 = 45 - 0 = 45^\circ \)C
Heat of fusion, \( L = 335 \times 10^3 \) J kg\(^{-1}\)
Coefficient of thermal conductivity of thermacole, \( K = 0.01 \) J s\(^{-1}\) m\(^{-1} \theta \) K\(^{-1}\)
Let \( m = \) mass (kg) of ice which melts,
Heat needed to melt at \( 0^\circ \)C, \( Q = mL \) .................... (1)
Also \( Q = KA \frac{\Delta\theta}{d} t \)
From (1) and (2), we get
\( m = \frac{KA\Delta\theta}{Ld} t \)
\( m = \frac{0.01 \times 0.54 \times 45}{335 \times 10^3 \times 5.0 \times 10^{-2}} \times 21600 \)
\( m = \frac{0.01 \times 0.54 \times 45 \times 21600}{335 \times 10^3 \times 0.05} \)
\( m = \frac{5248.8}{16750} \)
\( m = 0.313 \) kg
Mass of ice left in the box \( = M - m \)
\( = 4.0 - 0.313 \)
\( = 3.687 \) kg
\( \approx 3.7 \) kg.
In simple words: We calculated the heat flowing into the icebox from the warmer outside over 6 hours, using the box's size, material, and temperature difference. Then, we used the heat of fusion for ice to find out how much ice would melt from that heat. Finally, we subtracted the melted ice from the initial amount to see how much ice was left.

Exam Tip: Pay close attention to unit consistency; convert all measurements to a single system (e.g., SI units) before performing calculations. Remember the formula for heat transfer through conduction and the heat required for a phase change.

 

Question 20. A brass boiler has a base area 0.15 m² and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s-1 m-1 K-1. (Heat of vaporization of water = 2256 × 103 J kg-1)
Answer:
Here, \( K = 109 \) J s\(^{-1}\) m\(^{-1}\) K\(^{-1}\)
\( A = 0.15 \) m\(^2\)
\( d = 1.0 \) cm \( = 1.0 \times 10^{-2} \) m
\( \theta_2 = 100^\circ \)C (boiling point of water)
Let \( \theta_1 = \) temperature of the part of the boiler in contact with the stove.
If Q be the amount of heat flowing per second through the base of the boiler, then
Boiler, then \( Q = \frac{KA(\theta_1 - \theta_2)}{d} \)
\( Q = \frac{109 \times 0.15 \times (\theta_1 - 100)}{1.0 \times 10^{-2}} \)
\( Q = 1635 (\theta_1 - 100) \) J s\(^{-1}\) .................... (1)
Also heat of vaporization of water,
\( L = 2256 \times 10^3 \) J kg\(^{-1}\)
Rate of boiling of water in the boiler,
\( \frac{m}{t} = \frac{6.0}{60} = 0.1 \) kg s\(^{-1}\)
Heat received by water per second, \( Q = ML \)
\( Q = 0.1 \times 2256 \times 10^3 \)
\( Q = 225.6 \times 10^3 \) J s\(^{-1}\) .................... (2)
From (1) and (2), we get
\( 1635 (\theta_1 - 100) = 225.6 \times 10^3 \)
\( \theta_1 - 100 = \frac{225.6 \times 10^3}{1635} \)
\( \theta_1 - 100 = 138 \)
\( \theta_1 = 100 + 138 \)
\( \theta_1 = 238^\circ \)C
In simple words: We used the rate at which water boils to calculate the amount of heat transferred through the boiler per second. Then, using the brass boiler's properties and the heat transfer formula, we worked backwards to find the temperature of the flame in contact with the boiler.

Exam Tip: Be careful with units, especially when dealing with rate (kg/min vs. kg/s). Ensure you correctly apply the formula for heat conduction and relate the heat transfer to the latent heat of vaporization for the boiling process.

 

Question 21. Explain why:
(a) a body with large reflectivity is a poor emitter.
(b) a brass tumbler feels much colder than wooden tray on a chilly day.
(c) an optical pyrfp|eter (for measuring high temperatures) calibrated for Di ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace.
(d) the Earth without its atmosphere would be inhospitably cold.
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.
Answer:
(a) We know that \( a + r + t = 1 \), where a, r, and t represent absorptance, reflectance, and transmittance respectively of the surface of the body. Here, t is also called emittance (e). According to Kirchhoff's law, \( e \propto a \), which means good absorbers are good emitters. Hence, poor reflectors are good emitters, and vice-versa. If 'r' (reflectance) is high, then 'a' (absorptance) is lower, and therefore 'e' (emittance) is also lower, making it a poor emitter.
(b) Brass has high thermal conductivity, meaning it is a good heat conductor. So, when you touch a brass tumbler, heat quickly moves from your body to the tumbler, making it feel colder. On the other hand, wood is a poor heat conductor, so heat does not move quickly from your body to a wooden tray, making it feel comparatively warmer.
(c) Let T be the temperature of the hot iron in the furnace. According to Stefan's law, the heat radiated per second per unit area (E) is given by \( E = \sigma T^4 \). When the body is placed in the open at temperature To, the heat radiated per second per unit area (E') is given by \( E' = \sigma (T^4 - T_0^4) \). Clearly, \( E' < E \), so the optical pyrometer shows a lower value for the temperature of a red hot iron piece in the open. It gives a correct value when the piece is in the furnace because the furnace walls radiate heat back, approximating black body conditions.
(d) Gases are generally insulators. Earth's atmosphere acts as an insulating blanket that helps trap heat, preventing it from escaping into space and reflecting some of it back to Earth. If the atmosphere were absent, Earth would naturally be colder as all its heat would escape.
(e) Steam has a much higher heat capacity (540 cal g\(^{-1}\)) compared to the heat capacity of water (80 cal g\(^{-1}\)) at the same temperature. Therefore, steam will warm a building more effectively than hot water because it releases significantly more heat upon condensation.
In simple words: Reflective things don't give off much heat. Brass feels colder than wood because it takes heat from your hand faster. A pyrometer reads lower in the open because heat escapes, but correctly in a furnace. Earth's atmosphere keeps it warm by trapping heat. Steam heats better than hot water because it carries much more heat energy.

Exam Tip: For these types of explanatory questions, link physical properties (like reflectivity, conductivity, heat capacity) directly to the observed phenomenon. Ensure your explanations are clear, concise, and scientifically accurate, citing relevant laws or principles where applicable.

 

Question 22. A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20°C?
Answer:
According to Newton's law of cooling, the rate of cooling is proportional to the difference in temperature.
Here, average of 80°C and 50°C \( = \frac{80+50}{2} = 65^\circ \)C
Temperature of surroundings \( = 20^\circ \)C
Difference \( = 65 - 20 = 45^\circ \)C
Under these conditions, the body cools \( 30^\circ \)C (from 80°C to 50°C) in 5 minutes.
Using Newton's Law of Cooling, \( \frac{\text{Change in temp.}}{\text{Time}} = k \times \text{Average Temp. Difference} \)
\( \frac{30}{5} = k \times 45^\circ \) .................... (1)
The average of 60°C and 30°C \( = \frac{60+30}{2} = 45^\circ \)C, which is \( 25^\circ \)C (45 – 20) above the room temperature. The body cools by \( 30^\circ \)C (from 60°C to 30°C) in a time t (say).
\( \frac{30}{t} = k \times 25 \) .................... (2)
Where k is the same for both situations.
Dividing (1) by (2) gives,
\( \frac{30/5}{30/t} = \frac{k \times 45}{k \times 25} \)
\( \frac{t}{5} = \frac{45}{25} \)
\( \frac{t}{5} = \frac{9}{5} \)
\( t = 9 \) min.
In simple words: We used Newton's Law of Cooling to find the cooling constant 'k' from the first scenario (80°C to 50°C). Then, we applied this constant to the second scenario (60°C to 30°C) to calculate the new time it takes for the body to cool, considering the same room temperature.

Exam Tip: For Newton's Law of Cooling problems, remember that the "difference in temperature" refers to the difference between the average temperature of the object during cooling and the constant temperature of the surroundings. Ensure consistency in units for time and temperature.