Class 11 Physics MCQ – Thermal Properties of Matter – Heat Transfer

This set of Class 11 Physics Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Thermal Properties of Matter – Heat Transfer”.

1. When water is kept in a very hot pan, the entire water heats up. Which mode of heat transfer is responsible for this?
a) Conduction
b) Convection
c) Radiation
d) Convection along with radiation
View Answer

Answer: b
Explanation: Conduction involves transfer of heat energy due to vibration of molecules at their positions, convection involves transfer of heat energy by the motion of molecules themselves and radiation involves heat transfer by electromagnetic waves. When water is placed in a very hot pan, the molecules of water in immediate contact with the pan get heated and move away allowing the colder molecules to settle down hence allowing them to touch the pan and get heated. This movement of heated molecules corresponds to convection.

2. Quantitatively heat conduction can be described as?
a) Heat flow for a given temperature difference
b) Time rate of heat flow for a given temperature difference
c) Temperature difference caused due to heat flow per unit time
d) Temperature difference per unit quantity of heat transferred
View Answer

Answer: b
Explanation: Heat transfer by conduction occurs due to an initial temperature difference. Heat flow per unit time for a given temperature difference can be a quantitative description of heat flow by conduction. Note that temperature difference causes heat flow and not the other way around.

3. A steel and a copper bar are joined end to end. The area of the steel bar is half the area of the copper bar. Their lengths are equal to 10cm each. Temperature of free ends of steel & copper rod are 0°C & 100°C respectively. KS = 50 J/smK & KC =384 J/smK. Calculate equivalent thermal conductivity of the combined bar.
a) 63°C
b) 63K
c) 336°C
d) 36K
View Answer

Answer: a
Explanation: Temperature at the fixed ends of both rods will be the same.
KS AS (T-0)/LS = KC AC (100-T)/ LC
∴ 50*1/2 LCT = 384* C (100-T)
⇒ 25T = 38400 – 384T
⇒ T = 63°C.

4. What is the expression for heat current? Q is heat, t is time, T is temperature.
a) ΔQ/Δt
b) ΔQ/ΔT
c) ΔT/Δt
d) ΔQ/ΔTΔt
View Answer

Answer: a
Explanation: Heat current is the amount of heat crossing per unit time through any cross section. So, its expression can be written as ΔQ/Δt.

5. Select the correct statement.
a) Radiation doesn’t require any medium
b) Convection involves heat transfer by vibration of particles at their position
c) Conduction involves movement of particles to transfer heat
d) Radiation is responsible for heat of gas flame heating the cooking pan
View Answer

Answer: a
Explanation: Radiation doesn’t require any medium for transfer of heat. When a cooking pan is heated over a gas stove, conduction is the process responsible for transfer of heat from flame to pan. Radiation involves electromagnetic waves transferring energy called radiant energy.
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6. At a given temperature a body emits energy of a single wavelength corresponding to that temperature. True or False?
a) True
b) False
View Answer

Answer: b
Explanation: Thermal radiation from a body shows a continuous spectrum of wavelengths. But at a given temperature there is a particular short range of wavelengths which contributes the maximum to the total radiant energy. The graph given below shows the intensity of different wavelengths at different temperatures.

7. What is the value of Wien’s constant?
a) 2.9*10-3mK
b) 2.9*10-4mK
c) 2.9*10-3cmK
d) 2.9*10-4cmK
View Answer

Answer: a
Explanation: Wein’s displacement law states that the product of temperature and wavelength contributing the most to energy is a constant equal to 2.9*10-3mK.

8. The electromagnetic energy emitted by a perfect radiator corresponding to a temperature T and surface area A is?
a) AσT4Δt, in unit time Δt
b) AeσT4Δt, in unit time Δt
c) Ae4 c) AeσT4
d) AσT4
View Answer

Answer: a
Explanation: The relation of energy is given by Stefan-Boltzman law: Energy per unit time = AσT4, where σ is called the Stefan-Boltzman constant. ∴ Energy = AσT4Δt, in unit time Δt.

9. What is the net amount of energy per unit time being received by a body having a temperature of 40 °C in a surrounding of 60°C? (The surface area of body = 2m2, σ = 5.67*10-8W/ m2K4 & emissivity of 0.4).
a) 122.4W
b) 0.47W
c) 33W
d) 86W
View Answer

Answer: a
Explanation: The body will receive some energy from the surrounding and also lose some to it.
The net energy received will be
= AeσT4 – AeσTS4
= 2*0.4*5.67*10-8(3334 – 3134)
= 122.4 W.

10. The sun emits light having maximum intensity at a wavelength 468nm. Assume emissivity of sun is 1. Calculate its surface temperature. σ= 5.67*10-8W/m2K4.
a) 6800K
b) 3400K
c) 6196K
d) 7200K
View Answer

Answer: c
Explanation: Using Wein’s displacement law: λT = 2.9*10-3,
we get T = 2.9*10-3/(468*10-9)
= 6196 K.

11. If we hold a metal rod from one end and dip the other in a source of fire, what will happen to the temperature of the rod at steady state?
a) Be uniform
b) Decrease with time
c) Remain constant at a single value throughout the length of the rod
d) Stay nonuniform
View Answer

Answer: d
Explanation: When a rod is held with one end in fire. That end will have a temperature equal to that of fire. The temperature will gradually decrease towards the other end. At steady state the rate of heat flow will be constant and temperature at each point will be constant, but will be varying non-uniformly through the length of the rod.

12. Two spheres, A & B, are made up of the same material. The radius of A is twice that of B, while the temperature of A is half that of B. What is their ratio of energy emitted per unit time (EA : EB)?
a) 1
b) 1/4
c) 1/2
d) 1/16
View Answer

Answer: b
Explanation: Energy emitted per unit time is given by Ae.
e & σ will be the same for both.
A = 4πr2
∴ EA : EB= rA2*TA 2/rB2*TB4
= 4*1 / 1*16 = 1/4.

Sanfoundry Global Education & Learning Series – Physics – Class 11.

To practice all chapters and topics of class 11 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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