Chapter 11: Thermal Properties of Matter
Physics XI : Complete NCERT Exercise Solutions
Practice these examination-oriented questions to master the concepts of rectilinear motion, velocity, acceleration, and kinematic equations. Each question includes a detailed solution to enhance your understanding.
Question 11.1
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Answer & Explanation:
Given: Triple point of neon = 24.57 K, Triple point of CO2 = 216.55 K
Conversion formulas:
• Celsius: \( T_C = T_K – 273.15 \)
• Fahrenheit: \( T_F = \frac{9}{5} T_C + 32 \)
For Neon:
\( T_C = 24.57 – 273.15 = \mathbf{-248.58 \, ^\circ C} \)
\( T_F = \frac{9}{5} \times (-248.58) + 32 = \mathbf{-415.44 \, ^\circ F} \)
For CO2:
\( T_C = 216.55 – 273.15 = \mathbf{-56.6 \, ^\circ C} \)
\( T_F = \frac{9}{5} \times (-56.6) + 32 = \mathbf{-69.88 \, ^\circ F} \)
Conversion formulas:
• Celsius: \( T_C = T_K – 273.15 \)
• Fahrenheit: \( T_F = \frac{9}{5} T_C + 32 \)
For Neon:
\( T_C = 24.57 – 273.15 = \mathbf{-248.58 \, ^\circ C} \)
\( T_F = \frac{9}{5} \times (-248.58) + 32 = \mathbf{-415.44 \, ^\circ F} \)
For CO2:
\( T_C = 216.55 – 273.15 = \mathbf{-56.6 \, ^\circ C} \)
\( T_F = \frac{9}{5} \times (-56.6) + 32 = \mathbf{-69.88 \, ^\circ F} \)
Question 11.2
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 & Explanation:
Triple point of water on Kelvin scale = 273.16 K.
Since both scales are absolute, they are linearly related: \( T \propto \text{scale reading} \).
For scale A: \( 200 \, A = 273.16 \, K \)
For scale B: \( 350 \, B = 273.16 \, K \)
Thus, \( \frac{T_A}{200} = \frac{T_B}{350} = \frac{T_K}{273.16} \)
Therefore: \( \mathbf{T_A = \frac{200}{350} T_B = \frac{4}{7} T_B} \)
Since both scales are absolute, they are linearly related: \( T \propto \text{scale reading} \).
For scale A: \( 200 \, A = 273.16 \, K \)
For scale B: \( 350 \, B = 273.16 \, K \)
Thus, \( \frac{T_A}{200} = \frac{T_B}{350} = \frac{T_K}{273.16} \)
Therefore: \( \mathbf{T_A = \frac{200}{350} T_B = \frac{4}{7} T_B} \)
Question 11.3
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law: \( R = R_0[1 + \alpha(T – T_0)] \). The resistance is 101.6 Ω at the triple point of water 273.16 K, 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 & Explanation:
\( R_0 = 101.6 \, \Omega \) at \( T_0 = 273.16 \, K \)
At \( T = 600.5 \, K \), \( R = 165.5 \, \Omega \)
Using \( R = R_0[1 + \alpha(T – T_0)] \):
\( 165.5 = 101.6[1 + \alpha(600.5 – 273.16)] \)
\( \alpha = \frac{165.5/101.6 – 1}{327.34} \approx 0.00192 \, K^{-1} \)
For \( R = 123.4 \, \Omega \):
\( 123.4 = 101.6[1 + 0.00192(T – 273.16)] \)
\( T \approx 273.16 + \frac{123.4/101.6 – 1}{0.00192} \)
\( \mathbf{T \approx 384.8 \, K} \)
At \( T = 600.5 \, K \), \( R = 165.5 \, \Omega \)
Using \( R = R_0[1 + \alpha(T – T_0)] \):
\( 165.5 = 101.6[1 + \alpha(600.5 – 273.16)] \)
\( \alpha = \frac{165.5/101.6 – 1}{327.34} \approx 0.00192 \, K^{-1} \)
For \( R = 123.4 \, \Omega \):
\( 123.4 = 101.6[1 + 0.00192(T – 273.16)] \)
\( T \approx 273.16 + \frac{123.4/101.6 – 1}{0.00192} \)
\( \mathbf{T \approx 384.8 \, K} \)
Question 11.4
Answer the following:
(a) Why is the triple point of water a better fixed point than melting/boiling points?
(b) What is the other fixed point on the Kelvin scale?
(c) Why is 273.15 used in \( T_C = T – 273.15 \) instead of 273.16?
(d) What is the triple point of water on an absolute scale with Fahrenheit-sized units?
(a) Why is the triple point of water a better fixed point than melting/boiling points?
(b) What is the other fixed point on the Kelvin scale?
(c) Why is 273.15 used in \( T_C = T – 273.15 \) instead of 273.16?
(d) What is the triple point of water on an absolute scale with Fahrenheit-sized units?
Answer & Explanation:
(a) Triple point is unique and reproducible; melting/boiling points vary with pressure.
(b) Absolute zero (0 K) is the other fixed point.
(c) 273.15 is used because 0°C (ice point) is 273.15 K, not 273.16 K (triple point).
(d) On Fahrenheit scale, between ice and steam points: 180 divisions. On absolute scale with same unit size: triple point = \( 273.16 \times \frac{9}{5} = \mathbf{491.69} \) (on that scale).
(b) Absolute zero (0 K) is the other fixed point.
(c) 273.15 is used because 0°C (ice point) is 273.15 K, not 273.16 K (triple point).
(d) On Fahrenheit scale, between ice and steam points: 180 divisions. On absolute scale with same unit size: triple point = \( 273.16 \times \frac{9}{5} = \mathbf{491.69} \) (on that scale).
Question 11.5
Two ideal gas thermometers A (oxygen) and B (hydrogen) give the following readings at triple point and normal melting point of sulphur. Find absolute temperature of sulphur as per each and explain the difference.
Answer & Explanation:
For an ideal gas at constant volume: \( P \propto T \)
\( T = \frac{P}{P_{\text{tp}}} \times 273.16 \)
Thermometer A (O2):
\( T_S = \frac{1.797 \times 10^5}{1.250 \times 10^5} \times 273.16 \approx 392.7 \, K \)
Thermometer B (H2):
\( T_S = \frac{0.287 \times 10^5}{0.200 \times 10^5} \times 273.16 \approx 391.6 \, K \)
Difference arises because real gases deviate from ideal behavior. To reduce discrepancy, use lower gas pressures.
\( T = \frac{P}{P_{\text{tp}}} \times 273.16 \)
Thermometer A (O2):
\( T_S = \frac{1.797 \times 10^5}{1.250 \times 10^5} \times 273.16 \approx 392.7 \, K \)
Thermometer B (H2):
\( T_S = \frac{0.287 \times 10^5}{0.200 \times 10^5} \times 273.16 \approx 391.6 \, K \)
Difference arises because real gases deviate from ideal behavior. To reduce discrepancy, use lower gas pressures.
Question 11.6
A steel tape calibrated at 27°C measures a rod as 63.0 cm at 45°C. What is the actual rod length at 45°C and at 27°C? (\( \alpha_{\text{steel}} = 1.20 \times 10^{-5} \, K^{-1} \))
Answer & Explanation:
At 45°C, the tape expands, so it under-measures.
Actual length = measured length \( \times [1 + \alpha(45 – 27)] \)
\( = 63.0 \times [1 + 1.20 \times 10^{-5} \times 18] \, \text{cm} \)
\( \approx 63.0136 \, \text{cm at 45°C} \)
At 27°C, the rod contracts:
Length at 27°C = \( \frac{63.0136}{[1 + 1.20 \times 10^{-5} \times 18]} \approx 63.0 \, \text{cm} \)
Actual length = measured length \( \times [1 + \alpha(45 – 27)] \)
\( = 63.0 \times [1 + 1.20 \times 10^{-5} \times 18] \, \text{cm} \)
\( \approx 63.0136 \, \text{cm at 45°C} \)
At 27°C, the rod contracts:
Length at 27°C = \( \frac{63.0136}{[1 + 1.20 \times 10^{-5} \times 18]} \approx 63.0 \, \text{cm} \)
Question 11.7
A steel wheel (hole dia 8.69 cm) is to be fitted on a shaft (dia 8.70 cm) at 27°C. To what temperature must the shaft be cooled for the wheel to slip on? (\( \alpha_{\text{steel}} = 1.20 \times 10^{-5} \, K^{-1} \))
Answer & Explanation:
Shaft must contract to diameter ≤ 8.69 cm.
\( \Delta D = 8.69 – 8.70 = –0.01 \, \text{cm} \)
\( \Delta T = \frac{\Delta D}{D \times \alpha} = \frac{-0.01}{8.70 \times 1.20 \times 10^{-5}} \approx –95.8 \, K \)
\( T_{\text{final}} = 27 – 95.8 = \mathbf{–68.8 \, ^\circ C} \)
The shaft must be cooled to about –69°C using dry ice.
\( \Delta D = 8.69 – 8.70 = –0.01 \, \text{cm} \)
\( \Delta T = \frac{\Delta D}{D \times \alpha} = \frac{-0.01}{8.70 \times 1.20 \times 10^{-5}} \approx –95.8 \, K \)
\( T_{\text{final}} = 27 – 95.8 = \mathbf{–68.8 \, ^\circ C} \)
The shaft must be cooled to about –69°C using dry ice.
Question 11.8
A hole in a copper sheet has diameter 4.24 cm at 27°C. What is its diameter at 227°C? (\( \alpha_{\text{Cu}} = 1.70 \times 10^{-5} \, K^{-1} \))
Answer & Explanation:
A hole expands like the material it is in.
\( \Delta T = 227 – 27 = 200 \, ^\circ C \)
\( \Delta D = D \alpha \Delta T = 4.24 \times 1.70 \times 10^{-5} \times 200 \)
\( \Delta D \approx 0.0144 \, \text{cm} \)
New diameter = 4.24 + 0.0144 = \( \mathbf{4.2544 \, \text{cm}} \)
\( \Delta T = 227 – 27 = 200 \, ^\circ C \)
\( \Delta D = D \alpha \Delta T = 4.24 \times 1.70 \times 10^{-5} \times 200 \)
\( \Delta D \approx 0.0144 \, \text{cm} \)
New diameter = 4.24 + 0.0144 = \( \mathbf{4.2544 \, \text{cm}} \)
Question 11.9
A brass wire (length 1.8 m, dia 2.0 mm) at 27°C is held taut between rigid supports. If cooled to –39°C, what tension develops? (\( \alpha_{\text{brass}} = 2.0 \times 10^{-5} \, K^{-1} \), \( Y = 0.91 \times 10^{11} \, \text{Pa} \))
Answer & Explanation:
Thermal stress if expansion is prevented: \( \text{Stress} = Y \alpha \Delta T \)
\( \Delta T = –39 – 27 = –66 \, ^\circ C \) (cooling)
\( \text{Stress} = 0.91 \times 10^{11} \times 2.0 \times 10^{-5} \times 66 \approx 1.2012 \times 10^8 \, \text{Pa} \)
Area \( A = \pi (d/2)^2 = \pi (1 \times 10^{-3})^2 \approx 3.14 \times 10^{-6} \, \text{m}^2 \)
Tension \( F = \text{Stress} \times A \approx 1.2012 \times 10^8 \times 3.14 \times 10^{-6} \approx \mathbf{377 \, N} \)
\( \Delta T = –39 – 27 = –66 \, ^\circ C \) (cooling)
\( \text{Stress} = 0.91 \times 10^{11} \times 2.0 \times 10^{-5} \times 66 \approx 1.2012 \times 10^8 \, \text{Pa} \)
Area \( A = \pi (d/2)^2 = \pi (1 \times 10^{-3})^2 \approx 3.14 \times 10^{-6} \, \text{m}^2 \)
Tension \( F = \text{Stress} \times A \approx 1.2012 \times 10^8 \times 3.14 \times 10^{-6} \approx \mathbf{377 \, N} \)
Question 11.10
A brass rod and a steel rod, each 50 cm long and 3.0 mm diameter, are joined end to end. The combination is heated from 40°C to 250°C. Find the change in length. Is thermal stress developed at the junction? (\( \alpha_{\text{brass}} = 2.0 \times 10^{-5} \, K^{-1} \), \( \alpha_{\text{steel}} = 1.2 \times 10^{-5} \, K^{-1} \))
Answer & Explanation:
\( \Delta T = 250 – 40 = 210 \, ^\circ C \)
\( \Delta L_{\text{brass}} = L \alpha_{\text{brass}} \Delta T = 0.5 \times 2.0 \times 10^{-5} \times 210 = 0.0021 \, \text{m} = 2.1 \, \text{mm} \)
\( \Delta L_{\text{steel}} = 0.5 \times 1.2 \times 10^{-5} \times 210 = 0.00126 \, \text{m} = 1.26 \, \text{mm} \)
Total expansion = 2.1 + 1.26 = \( \mathbf{3.36 \, \text{mm}} \)
Since ends are free, no thermal stress is developed at the junction.
\( \Delta L_{\text{brass}} = L \alpha_{\text{brass}} \Delta T = 0.5 \times 2.0 \times 10^{-5} \times 210 = 0.0021 \, \text{m} = 2.1 \, \text{mm} \)
\( \Delta L_{\text{steel}} = 0.5 \times 1.2 \times 10^{-5} \times 210 = 0.00126 \, \text{m} = 1.26 \, \text{mm} \)
Total expansion = 2.1 + 1.26 = \( \mathbf{3.36 \, \text{mm}} \)
Since ends are free, no thermal stress is developed at the junction.
Question 11.11
The coefficient of volume expansion of glycerine is \( 49 \times 10^{-5} \, K^{-1} \). Find the fractional change in density for a 30°C rise in temperature.
Answer & Explanation:
For a substance: \( \rho \propto 1/V \)
Fractional change in density: \( \frac{\Delta \rho}{\rho} = – \frac{\Delta V}{V} = – \alpha_V \Delta T \)
\( \frac{\Delta \rho}{\rho} = – 49 \times 10^{-5} \times 30 = \mathbf{–0.0147} \)
Density decreases by about 1.47%.
Fractional change in density: \( \frac{\Delta \rho}{\rho} = – \frac{\Delta V}{V} = – \alpha_V \Delta T \)
\( \frac{\Delta \rho}{\rho} = – 49 \times 10^{-5} \times 30 = \mathbf{–0.0147} \)
Density decreases by about 1.47%.
Question 11.12
A 10 kW drilling machine drills into an 8.0 kg aluminium block for 2.5 minutes. If 50% power is used to heat the block, find its temperature rise. (\( s_{\text{Al}} = 0.91 \, \text{J g}^{-1} \text{K}^{-1} \))
Answer & Explanation:
Power used for heating = 50% of 10 kW = 5000 W
Time = 2.5 min = 150 s
Heat supplied \( Q = 5000 \times 150 = 7.5 \times 10^5 \, \text{J} \)
\( Q = m s \Delta T \)
\( \Delta T = \frac{Q}{m s} = \frac{7.5 \times 10^5}{8.0 \times 0.91 \times 10^3} \)
\( \Delta T \approx \mathbf{103 \, ^\circ C} \)
Time = 2.5 min = 150 s
Heat supplied \( Q = 5000 \times 150 = 7.5 \times 10^5 \, \text{J} \)
\( Q = m s \Delta T \)
\( \Delta T = \frac{Q}{m s} = \frac{7.5 \times 10^5}{8.0 \times 0.91 \times 10^3} \)
\( \Delta T \approx \mathbf{103 \, ^\circ C} \)
Question 11.13
A 2.5 kg copper block at 500°C is placed on a large ice block. What maximum mass of ice can melt? (\( s_{\text{Cu}} = 0.39 \, \text{J g}^{-1} \text{K}^{-1} \), \( L_f = 335 \, \text{J g}^{-1} \))
Answer & Explanation:
Heat lost by copper = \( m_{\text{Cu}} s_{\text{Cu}} \Delta T \)
\( = 2500 \times 0.39 \times 500 = 4.875 \times 10^5 \, \text{J} \)
This heat melts ice: \( Q = m_{\text{ice}} L_f \)
\( m_{\text{ice}} = \frac{Q}{L_f} = \frac{4.875 \times 10^5}{335} \approx \mathbf{1455 \, g \approx 1.46 \, kg} \)
\( = 2500 \times 0.39 \times 500 = 4.875 \times 10^5 \, \text{J} \)
This heat melts ice: \( Q = m_{\text{ice}} L_f \)
\( m_{\text{ice}} = \frac{Q}{L_f} = \frac{4.875 \times 10^5}{335} \approx \mathbf{1455 \, g \approx 1.46 \, kg} \)
Question 11.14
A 0.20 kg metal block at 150°C is dropped into a calorimeter with 150 cm³ water at 27°C. Final temperature is 40°C. Find specific heat of metal. Will heat loss make your value higher or lower?
Answer & Explanation:
Heat lost by metal = heat gained by (water + calorimeter)
Water equivalent of calorimeter = 0.025 kg
\( m_w = 0.15 \, \text{kg} \) (since 150 cm³ water ≈ 0.15 kg)
\( 0.20 \times s \times (150 – 40) = (0.15 + 0.025) \times 4186 \times (40 – 27) \)
\( s \approx 434 \, \text{J kg}^{-1} \text{K}^{-1} \)
If heat is lost to surroundings, calculated s will be smaller than actual.
Water equivalent of calorimeter = 0.025 kg
\( m_w = 0.15 \, \text{kg} \) (since 150 cm³ water ≈ 0.15 kg)
\( 0.20 \times s \times (150 – 40) = (0.15 + 0.025) \times 4186 \times (40 – 27) \)
\( s \approx 434 \, \text{J kg}^{-1} \text{K}^{-1} \)
If heat is lost to surroundings, calculated s will be smaller than actual.
Question 11.15
Why do diatomic gases have higher molar specific heat (~5 cal/mol·K) than monatomic gases (~3 cal/mol·K)? Why is chlorine’s value higher?
Answer & Explanation:
Monatomic gases have only translational degrees of freedom (3), giving \( C_V = \frac{3}{2} R \approx 3 \, \text{cal/mol·K} \).
Diatomic gases have 3 translational + 2 rotational (at room temp) = 5 degrees of freedom, so \( C_V = \frac{5}{2} R \approx 5 \, \text{cal/mol·K} \).
Chlorine has a slightly larger value because at room temperature, vibrational modes begin to contribute slightly.
Diatomic gases have 3 translational + 2 rotational (at room temp) = 5 degrees of freedom, so \( C_V = \frac{5}{2} R \approx 5 \, \text{cal/mol·K} \).
Chlorine has a slightly larger value because at room temperature, vibrational modes begin to contribute slightly.
Question 11.16
A child (30 kg) with fever 101°F is given medicine. Fever reduces to 98°F in 20 min by sweat evaporation. Find average extra evaporation rate. (\( s_{\text{body}} \approx s_{\text{water}} \), \( L = 580 \, \text{cal/g} \))
Answer & Explanation:
\( \Delta T \) in °C: \( (101 – 98) \times \frac{5}{9} \approx 1.67 \, ^\circ C \)
Heat to be lost \( Q = m s \Delta T = 30000 \times 1 \times 1.67 \approx 50100 \, \text{cal} \)
This heat is removed by evaporation in 20 min = 1200 s
Evaporation rate = \( \frac{Q}{L \times t} = \frac{50100}{580 \times 1200} \approx 0.072 \, \text{g/s} \)
\( \approx \mathbf{4.32 \, \text{g/min}} \)
Heat to be lost \( Q = m s \Delta T = 30000 \times 1 \times 1.67 \approx 50100 \, \text{cal} \)
This heat is removed by evaporation in 20 min = 1200 s
Evaporation rate = \( \frac{Q}{L \times t} = \frac{50100}{580 \times 1200} \approx 0.072 \, \text{g/s} \)
\( \approx \mathbf{4.32 \, \text{g/min}} \)
Question 11.17
A thermacole icebox (side 30 cm, thickness 5 cm) contains 4.0 kg ice at 0°C. Outside temperature is 45°C. Estimate ice remaining after 6 h. (\( K = 0.01 \, \text{J s}^{-1} \text{m}^{-1} \text{K}^{-1} \), \( L_f = 335 \times 10^3 \, \text{J/kg} \))
Answer & Explanation:
Surface area \( A = 6 \times (0.30)^2 = 0.54 \, \text{m}^2 \)
Thickness \( d = 0.05 \, \text{m} \), \( \Delta T = 45 – 0 = 45 \, ^\circ C \)
Heat flow rate \( H = \frac{K A \Delta T}{d} \)
\( = \frac{0.01 \times 0.54 \times 45}{0.05} = 4.86 \, \text{J/s} \)
Total time = \( 6 \times 3600 = 21600 \, \text{s} \)
Total heat entering \( Q = 4.86 \times 21600 \approx 1.05 \times 10^5 \, \text{J} \)
Mass of ice melted = \( \frac{Q}{L_f} = \frac{1.05 \times 10^5}{335000} \approx 0.313 \, \text{kg} \)
Ice remaining = 4.0 – 0.313 ≈ \( \mathbf{3.687 \, \text{kg}} \)
Thickness \( d = 0.05 \, \text{m} \), \( \Delta T = 45 – 0 = 45 \, ^\circ C \)
Heat flow rate \( H = \frac{K A \Delta T}{d} \)
\( = \frac{0.01 \times 0.54 \times 45}{0.05} = 4.86 \, \text{J/s} \)
Total time = \( 6 \times 3600 = 21600 \, \text{s} \)
Total heat entering \( Q = 4.86 \times 21600 \approx 1.05 \times 10^5 \, \text{J} \)
Mass of ice melted = \( \frac{Q}{L_f} = \frac{1.05 \times 10^5}{335000} \approx 0.313 \, \text{kg} \)
Ice remaining = 4.0 – 0.313 ≈ \( \mathbf{3.687 \, \text{kg}} \)
Question 11.18
A brass boiler (base area 0.15 m², thickness 1 cm) boils water at 6 kg/min. Estimate flame temperature in contact. (\( K = 109 \, \text{J s}^{-1} \text{m}^{-1} \text{K}^{-1} \), \( L_v = 2256 \times 10^3 \, \text{J/kg} \))
Answer & Explanation:
Rate of boiling = 6 kg/min = 0.1 kg/s
Heat required per second = \( \dot{m} L_v = 0.1 \times 2256 \times 10^3 = 2.256 \times 10^4 \, \text{J/s} \)
This heat flows through base: \( H = \frac{K A \Delta T}{d} \)
\( 2.256 \times 10^4 = \frac{109 \times 0.15 \times \Delta T}{0.01} \)
\( \Delta T \approx 13.8 \, ^\circ C \)
Water boils at 100°C, so flame temperature ≈ 100 + 13.8 ≈ \( \mathbf{113.8 \, ^\circ C} \)
Heat required per second = \( \dot{m} L_v = 0.1 \times 2256 \times 10^3 = 2.256 \times 10^4 \, \text{J/s} \)
This heat flows through base: \( H = \frac{K A \Delta T}{d} \)
\( 2.256 \times 10^4 = \frac{109 \times 0.15 \times \Delta T}{0.01} \)
\( \Delta T \approx 13.8 \, ^\circ C \)
Water boils at 100°C, so flame temperature ≈ 100 + 13.8 ≈ \( \mathbf{113.8 \, ^\circ C} \)
Question 11.19
Explain: (a) Why good reflectors are poor emitters, (b) Why brass feels colder than wood, (c) Why pyrometer gives low reading for red hot iron in open, (d) Why Earth without atmosphere would be cold, (e) Why steam heating is more efficient than hot water.
Answer & Explanation:
(a) By Kirchhoff’s law, good reflectors absorb less radiation, hence emit less.
(b) Brass has higher thermal conductivity, so it draws heat quickly from hand, feeling colder.
(c) In open, iron loses heat by convection/radiation; pyrometer (calibrated for black body) sees lower effective radiation.
(d) Atmosphere traps IR radiation via greenhouse effect; without it, Earth radiates freely into space.
(e) Steam releases latent heat (2260 kJ/kg) in addition to sensible heat, transferring more energy per kg.
(b) Brass has higher thermal conductivity, so it draws heat quickly from hand, feeling colder.
(c) In open, iron loses heat by convection/radiation; pyrometer (calibrated for black body) sees lower effective radiation.
(d) Atmosphere traps IR radiation via greenhouse effect; without it, Earth radiates freely into space.
(e) Steam releases latent heat (2260 kJ/kg) in addition to sensible heat, transferring more energy per kg.
Question 11.20
A body cools from 80°C to 50°C in 5 minutes in surroundings at 20°C. Find time to cool from 60°C to 30°C.
Answer & Explanation:
For small ΔT, Newton’s law of cooling: \( \frac{dT}{dt} = –k(T – T_s) \)
Average temperature in first case: \( \frac{80+50}{2} = 65 \, ^\circ C \), excess = 45°C
Cooling rate = \( \frac{30 \, ^\circ C}{5 \, \text{min}} = 6 \, ^\circ C/\text{min} \propto 45 \, ^\circ C \) → \( k = \frac{6}{45} \, \text{per min} \)
In second case: average = \( \frac{60+30}{2} = 45 \, ^\circ C \), excess = 25°C
Rate ∝ 25°C → rate = \( \frac{6}{45} \times 25 = \frac{10}{3} \, ^\circ C/\text{min} \)
Time = \( \frac{\Delta T}{\text{rate}} = \frac{30}{10/3} = \mathbf{9 \, \text{minutes}} \)
Average temperature in first case: \( \frac{80+50}{2} = 65 \, ^\circ C \), excess = 45°C
Cooling rate = \( \frac{30 \, ^\circ C}{5 \, \text{min}} = 6 \, ^\circ C/\text{min} \propto 45 \, ^\circ C \) → \( k = \frac{6}{45} \, \text{per min} \)
In second case: average = \( \frac{60+30}{2} = 45 \, ^\circ C \), excess = 25°C
Rate ∝ 25°C → rate = \( \frac{6}{45} \times 25 = \frac{10}{3} \, ^\circ C/\text{min} \)
Time = \( \frac{\Delta T}{\text{rate}} = \frac{30}{10/3} = \mathbf{9 \, \text{minutes}} \)
Question 11.21
Based on CO2 phase diagram: (a) Triple point coordinates, (b) Effect of pressure decrease on fusion/boiling points, (c) Critical T & P, (d) State of CO2 at given conditions.
Answer & Explanation:
(a) Triple point: ~5.11 atm, –56.6°C
(b) Lower pressure lowers boiling point; fusion point slightly decreases.
(c) Critical point: ~73 atm, 31°C; beyond this, liquid and gas phases are indistinguishable.
(d) (i) –70°C, 1 atm: solid (sublimes), (ii) –60°C, 10 atm: liquid, (iii) 15°C, 56 atm: liquid.
(b) Lower pressure lowers boiling point; fusion point slightly decreases.
(c) Critical point: ~73 atm, 31°C; beyond this, liquid and gas phases are indistinguishable.
(d) (i) –70°C, 1 atm: solid (sublimes), (ii) –60°C, 10 atm: liquid, (iii) 15°C, 56 atm: liquid.
Question 11.22
Based on CO2 phase diagram: (a) Isothermal compression at 1 atm, –60°C, (b) Cooling at 4 atm from RT, (c) Heating solid CO2 at 10 atm from –65°C to RT, (d) Isothermal compression at 70°C.
Answer & Explanation:
(a) At –60°C and 1 atm, CO2 is vapor; compression leads directly to solid (no liquid phase).
(b) At 4 atm, cooling from RT: gas → liquid → solid.
(c) At 10 atm, –65°C: solid → liquid (at ~–40°C) → gas (at ~–25°C).
(d) At 70°C (above critical T), isothermal compression: gas → supercritical fluid (no phase change).
(b) At 4 atm, cooling from RT: gas → liquid → solid.
(c) At 10 atm, –65°C: solid → liquid (at ~–40°C) → gas (at ~–25°C).
(d) At 70°C (above critical T), isothermal compression: gas → supercritical fluid (no phase change).
📘 Exam Preparation Tip:
These exercise questions cover key concepts: temperature scales, thermal expansion, calorimetry, specific heat, change of state, heat transfer modes, Newton’s law of cooling, and phase diagrams. Solving these helps understand real-world applications like cooling systems, climate effects, and material behavior under temperature changes—essential for board and competitive exams.