JEE Physics Thermal Properties Of Matter MCQs Set 20

Physics Objective Questions and Answers: Thermal Properties Of Matter

Access targeted multiple-choice questions for Thermal Properties Of Matter designed to align with the latest JEE academic syllabus for JEE Physics. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.

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Multiple Choice Questions

Question 1: A container contains hot water at 100°C. If in time T₁ the temperature falls to 80°C and in time T₂ the temperature falls to 60°C from 80°C, then
(a) \( T_1=T_2 \)
(b) \( T_1 \gt T_2 \)
(c) \( T_1 \lt T_2 \)
(d) none of these
Show Answer & Explanation

Answer: (c) \( T_1 \lt T_2 \)

Explanation:
1. The cooling rate falls as the body approaches the surroundings.
2. The second fall (80 to 60°C) occurs at a lower average temperature, so it takes longer: T₁ < T₂.

Teacher's Note:
1. Cooling slows down.

Question 2: Which of the following will expand the most for the same rise in temperature?
(a) Aluminium
(b) Glass
(c) Wood
(d) All will expand the same
Show Answer & Explanation

Answer: (a) Aluminium

Explanation:
1. Aluminium has a linear expansion coefficient of about 2.3 × 10⁻⁵ /°C, much larger than glass (0.9 × 10⁻⁵) or wood.
2. So it expands the most.

Teacher's Note:
1. Metals expand more than glass.

Question 3: Four identical rods of the same material are joined end to end to form a square. If the temperature difference between the ends of a diagonal is 100°C, then the temperature difference between the ends of the other diagonal will be
(a) 0°C
(b) \( \frac{100}{l}\,^\circ\text{C} \)
(c) \( \frac{100}{2l}\,^\circ\text{C} \)
(d) 100°C
Show Answer & Explanation

Answer: (a) 0°C

Explanation:
1. The square is a balanced Wheatstone-bridge arrangement.
2. The other two corners are at the same temperature, so the difference is 0°C.

Teacher's Note:
1. Symmetry makes the other diagonal isothermal.

Question 4: Consider two hot bodies B₁ and B₂ which have temperatures 100°C and 80°C respectively at t = 0. The temperature of the surroundings is 40°C. The ratio of the respective rates of cooling R₁ and R₂ of these two bodies at t = 0 will be
(a) \( R_1:R_2=3:2 \)
(b) \( R_1:R_2=5:4 \)
(c) \( R_1:R_2=2:3 \)
(d) \( R_1:R_2=4:5 \)
Show Answer & Explanation

Answer: (a) \( R_1:R_2=3:2 \)

Explanation:
1. By Newton's law, the rate is proportional to the excess temperature.
2. R₁ : R₂ = (100 − 40) : (80 − 40) = 3 : 2.

Teacher's Note:
1. Rate is proportional to temperature excess.

Question 5: Assuming no heat losses, the heat released by the condensation of x g of steam at 100°C can be used to convert y g of ice at 0°C into water at 100°C. The ratio x : y is
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 3 : 1
Show Answer & Explanation

Answer: (c) 1 : 3

Explanation:
1. Steam condensing at 100°C releases 540x cal.
2. Ice melts and warms to 100°C: y(80 + 100) = 180y cal.
3. 540x = 180y, so x : y = 1 : 3.

Teacher's Note:
1. Equate heat released with heat absorbed.

Question 6: Newton's law of cooling is applicable for
(a) any excess of temperature over the surroundings
(b) a small excess of temperature over the surroundings
(c) a large excess of temperature over the surroundings
(d) a very large excess of temperature over the surroundings
Show Answer & Explanation

Answer: (b) a small excess of temperature over the surroundings

Explanation:
1. The law holds when the temperature excess is small, so that heat loss is mostly by convection and radiation linear in the excess.
2. Large excess needs Stefan's law.

Teacher's Note:
1. Small temperature excess.

Question 7: A lamp black surface absorbs radiant heat which is near about
(a) 90%
(b) 98%
(c) 100%
(d) 50%
Show Answer & Explanation

Answer: (b) 98%

Explanation:
1. Lamp black is very close to a perfect blackbody and absorbs about 98% (near 96–98%) of incident radiation.
2. It does not absorb 100%.

Teacher's Note:
1. Near-black body.

Question 8: 4200 J of work is required for
(a) increasing the temperature of 10 g of water through 10°C
(b) increasing the temperature of 100 g of water through 10°C
(c) increasing the temperature of 1 kg of water through 10°C
(d) increasing the temperature of 10 g of water through 100°C
Show Answer & Explanation

Answer: (b) increasing the temperature of 100 g of water through 10°C

Explanation:
1. Q = mcΔT with c = 4200 J/kg°C.
2. For 0.1 kg and 10°C, Q = 0.1 × 4200 × 10 = 4200 J.

Teacher's Note:
1. Use Q = mcΔT.

Question 9: Which of the following undergo sublimation?
(a) Dry ice
(b) Iodine
(c) Both (a) and (b)
(d) None of these
Show Answer & Explanation

Answer: (c) Both (a) and (b)

Explanation:
1. Both solid carbon dioxide and iodine change directly from solid to vapour.
2. So both sublime.

Teacher's Note:
1. Solid to gas directly.

Question 10: Consider the following statements and select the correct statement(s). I. Water and ice have the same specific heats. II. Water and ice have different specific heats. III. The specific heat of water is more than that of ice. IV. The specific heat of ice is more than that of water.
(a) I only
(b) II only
(c) II and III
(d) III and IV
Show Answer & Explanation

Answer: (c) II and III

Explanation:
1. Water has c = 4200 J/kg K and ice about 2100 J/kg K.
2. So they differ and water's is larger.

Teacher's Note:
1. c_water is about twice c_ice.

Question 11: Three liquids A, B and C having the same specific heat and masses m, 2m and 3m have temperatures 20°C, 40°C and 60°C respectively. Match the temperature of the mixture when
Column IColumn II
(A) A and B are mixed(1) 33.3°C
(B) A and C are mixed(2) 52°C
(C) B and C are mixed(3) 50°C
(D) A, B and C all three are mixed(4) 46.67°C

(a) A-2, B-3, C-1, D-4
(b) A-1, B-4, C-2, D-3
(c) A-3, B-1, C-2, D-4
(d) A-1, B-3, C-2, D-4
Show Answer & Explanation

Answer: (d) A-1, B-3, C-2, D-4

Explanation:
1. A and B: (20m + 80m)/3m = 33.3°C.
2. A and C: (20m + 180m)/4m = 50°C.
3. B and C: (80m + 180m)/5m = 52°C.
4. All three: (20 + 80 + 180)m/6m = 46.67°C.

Teacher's Note:
1. Weighted average of temperatures.

Question 12: A system of two concentric spheres of radii r₁ and r₂ are kept at temperatures T₁ and T₂, respectively. The radial rate of flow of heat in a substance between the two concentric spheres is proportional to
(a) \( \ln\left(\frac{r_2}{r_1}\right) \)
(b) \( \frac{(r_2-r_1)}{(r_1r_2)} \)
(c) \( (r_2-r_1) \)
(d) \( \frac{r_1r_2}{(r_2-r_1)} \)
Show Answer & Explanation

Answer: (d) \( \frac{r_1r_2}{(r_2-r_1)} \)

Explanation:
1. Integrating dQ/dt = −K 4πr² dT/dr gives dQ/dt = 4πK r₁r₂(T₁ − T₂)/(r₂ − r₁).
2. So the rate is proportional to r₁r₂/(r₂ − r₁).

Teacher's Note:
1. Spherical conduction formula.

Question 13:
Assertion (A): Two bodies at different temperatures, if brought in thermal contact, do not necessarily settle to the mean temperature.
Reason (R): The two bodies may have different thermal capacities.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The final temperature is the capacity-weighted average.
2. It equals the plain mean only when the thermal capacities are equal.

Teacher's Note:
1. T = (C₁T₁ + C₂T₂)/(C₁ + C₂).

Question 14: The two ends of a metal rod are maintained at temperatures 100°C and 110°C. The rate of heat flow in the rod is found to be 4.0 J/s. If the ends are maintained at temperatures 200°C and 210°C, the rate of heat flow will be
(a) 16.8 J/s
(b) 8.0 J/s
(c) 4.0 J/s
(d) 44.0 J/s
Show Answer & Explanation

Answer: (c) 4.0 J/s

Explanation:
1. Rate = KAΔT/L depends only on the temperature difference.
2. Both cases have ΔT = 10°C, so the rate is 4.0 J/s.

Teacher's Note:
1. Only ΔT matters.

Question 15: Which of the following pairs may give equal numerical values of the temperature of a body?
(a) Fahrenheit and Celsius
(b) Celsius and Kelvin
(c) Kelvin and Reaumur
(d) none of these
Show Answer & Explanation

Answer: (a) Fahrenheit and Celsius

Explanation:
1. T_C/100 = (T_F − 32)/180 gives T_C = T_F at −40.
2. Celsius and Kelvin readings differ by 273.15 and cannot be equal.

Teacher's Note:
1. The scales meet at −40.

Question 16: If α, β and γ are the coefficients of linear, area and volume expansion respectively, then (for an isotropic solid)
(a) \( \gamma=3\alpha \)
(b) \( \alpha=3\gamma \)
(c) \( \beta=3\alpha \)
(d) \( \gamma=3\beta \)
Show Answer & Explanation

Answer: (a) \( \gamma=3\alpha \)

Explanation:
1. Volume is length cubed, so γ = 3α.
2. Also β = 2α.

Teacher's Note:
1. γ : β : α = 3 : 2 : 1.

Question 17:
Assertion (A): The specific heat of a body is always greater than its thermal capacity.
Reason (R): Specific heat capacity is the heat required for raising the temperature of unit mass of the body through a unit degree.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (d) A is false, but R is true.

Explanation:
1. Thermal capacity = m × specific heat.
2. For a body of mass less than 1 unit it may not be greater, so the assertion is false while the reason is a correct definition.

Teacher's Note:
1. Thermal capacity equals mass times specific heat.

Question 18: Steam is passed into 22 g of water at 20°C. The mass of water that will be present when the water acquires a temperature of 90°C is (latent heat of steam = 540 cal/g)
(a) 24.8 g
(b) 24 g
(c) 36.6 g
(d) 30 g
Show Answer & Explanation

Answer: (a) 24.8 g

Explanation:
1. Heat gained by the water = 22 × 70 = 1540 cal.
2. Each gram of steam gives 540 + 10 = 550 cal, so the steam mass = 2.8 g.
3. Total water = 22 + 2.8 = 24.8 g.

Teacher's Note:
1. Steam condenses and then cools.

Question 19: A bar of iron is 10 cm at 20°C. At 19°C it will be (α of iron = 11 × 10⁻⁶ /°C)
(a) 11 × 10⁻⁶ cm longer
(b) 11 × 10⁻⁶ cm shorter
(c) 11 × 10⁻⁵ cm longer
(d) 11 × 10⁻⁵ cm shorter
Show Answer & Explanation

Answer: (d) 11 × 10⁻⁵ cm shorter

Explanation:
1. ΔL = Lα ΔT = 10 × 11 × 10⁻⁶ × 1 = 11 × 10⁻⁵ cm.
2. The temperature falls, so the bar is shorter.

Teacher's Note:
1. Cooling contracts the bar.

Question 20: Which of the following is an expression for heat capacity?
(a) \( S=\frac1m\frac{\Delta Q}{\Delta T} \)
(b) \( S=\frac{\Delta Q}{\Delta T} \)
(c) \( S=\frac1\mu\frac{\Delta Q}{\Delta T} \)
(d) all of the above
Show Answer & Explanation

Answer: (b) \( S=\frac{\Delta Q}{\Delta T} \)

Explanation:
1. Heat capacity is the heat needed to raise the temperature of the whole body by one degree.
2. S = ΔQ/ΔT; the other forms are specific heat and molar specific heat.

Teacher's Note:
1. Heat capacity is for the whole body.

Question 21: Three bodies A, B and C have equal area and are painted red, yellow and black respectively. If they are at the same temperature, then
(a) the emissive power of A is maximum
(b) the emissive power of B is maximum
(c) the emissive power of C is maximum
(d) the emissive power of A, B and C are equal
Show Answer & Explanation

Answer: (c) the emissive power of C is maximum

Explanation:
1. A black surface is the best absorber and hence the best emitter.
2. So body C (black) has the maximum emissive power.

Teacher's Note:
1. Black is the best emitter.

Question 22: The reading of a Centigrade thermometer coincides with that of a Fahrenheit thermometer in a liquid. The temperature of the liquid is
(a) −40°C
(b) 313°C
(c) 0°C
(d) 100°C
Show Answer & Explanation

Answer: (a) −40°C

Explanation:
1. Set T_C = T_F = T in T_C/5 = (T_F − 32)/9.
2. 9T = 5T − 160 gives T = −40.

Teacher's Note:
1. −40 is the same on both scales.

Question 23: Sweet makers do not clean the bottom of their cauldrons because
(a) the emission power of black and bright surfaces is more
(b) the absorption power of black and bright surfaces is more
(c) black and rough surfaces absorb more heat
(d) the transmission power of black and rough surfaces is more
Show Answer & Explanation

Answer: (c) black and rough surfaces absorb more heat

Explanation:
1. The sooty black rough bottom absorbs radiation and heat much better.
2. So the cauldron heats up faster and uses less fuel.

Teacher's Note:
1. Black rough surfaces absorb well.

Question 24: Three rods of the same dimensions have thermal conductivities 3K, 2K and K. They are joined at a common junction, with the free ends of the rods at 100°C (3K rod), 50°C (2K rod) and 20°C (K rod). The temperature of the junction is
(a) 60°C
(b) 70°C
(c) 50°C
(d) 35°C
Show Answer & Explanation

Answer: (b) 70°C

Explanation:
1. In steady state the heat entering the junction equals the heat leaving it.
2. 3K(100 − θ) = 2K(θ − 50) + K(θ − 20) gives 300 − 3θ = 3θ − 120, so θ = 70°C.

Teacher's Note:
1. Heat in equals heat out at the junction.

Question 25: Which of the following is used as a coolant in an automobile radiator as well as a heater in hot water bags?
(a) Ice
(b) Sand
(c) Water
(d) All of these
Show Answer & Explanation

Answer: (c) Water

Explanation:
1. Water has a high specific heat.
2. It absorbs or releases large amounts of heat with a small change in temperature.

Teacher's Note:
1. High specific heat.

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