Thermal Radiation — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Thermal Radiation MCQs with step-by-step solutions (31 questions). Part of Kinetic Theory of Gases & Radiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Thermal Radiation · easy · theory
A perfect blackbody is one that:
A. Transmits all the radiation incident upon it
B. Absorbs all the radiation incident upon it ✓ Correct
C. Emits no radiation at any temperature
D. Reflects all the radiation incident upon it
Solution: A perfect blackbody has absorptive power $a = 1$: it neither reflects nor transmits. Being the best absorber, it is also the best emitter at any given temperature.
Q2 — Thermal Radiation · easy · theory
The emissivity of a perfect blackbody is:
A. $0$
B. $0.5$
C. Infinite
D. $1$ ✓ Correct
Solution: Emissivity is defined relative to a blackbody, which by definition radiates the maximum possible at each temperature, so $e = 1$ for it and $0 < e < 1$ for real surfaces.
Q3 — Thermal Radiation · easy · theory
Kirchhoff's law of radiation states that, at a given temperature and wavelength:
A. Good absorbers of radiation are equally good emitters ✓ Correct
B. Good absorbers of radiation are poor emitters
C. Emission is independent of the nature of the surface
D. All bodies emit the same amount of radiation
Solution: Kirchhoff's law states that the ratio of emissive power to absorptive power is the same for all bodies and equals the emissive power of a blackbody, so $a$ and $e$ go together.
Q4 — Thermal Radiation · easy · theory
For radiation incident on a body, the absorptive power $a$, reflectance $r$ and transmittance $t$ satisfy:
A. $a + r = t$
B. $a + r + t = 1$ ✓ Correct
C. $a + r + t = 0$
D. $a = r = t$
Solution: Energy is conserved: whatever fraction of the incident radiation is not reflected or transmitted must be absorbed.
Q5 — Thermal Radiation · easy · theory
Heat transfer by radiation differs from conduction and convection in that radiation:
A. Travels more slowly than sound
B. Requires no material medium for its propagation ✓ Correct
C. Always requires a fluid medium
D. Occurs only in solids
Solution: Thermal radiation consists of electromagnetic waves, which travel through vacuum at the speed of light. That is how the Sun's energy reaches the Earth.
Q6 — Thermal Radiation · easy · theory
The coefficient of emission (emissivity) of a surface is:
A. Measured in $\text{J}/\text{K}$
B. A dimensionless number lying between $0$ and $1$ ✓ Correct
C. Measured in $\text{W}/\text{m}^2$
D. Always greater than $1$
Solution: It is the ratio of the emissive power of the surface to that of a perfect blackbody at the same temperature, so it is a pure fraction.
Q7 — Thermal Radiation · medium · theory
Prevost's theory of heat exchange states that a body:
A. Radiates energy only when it is hotter than its surroundings
B. Stops radiating once it reaches thermal equilibrium
C. Radiates energy at all temperatures above absolute zero ✓ Correct
D. Absorbs energy only when it is colder than its surroundings
Solution: Radiation and absorption go on continuously at every temperature above $0\text{ K}$. At equilibrium the two rates are simply equal, so there is no net exchange.
Q8 — Thermal Radiation · easy · theory
Newton's law of cooling states that the rate of loss of heat of a body is proportional to:
A. Its total heat content
B. The square of its absolute temperature
C. The fourth power of its absolute temperature
D. The excess of its temperature over that of the surroundings ✓ Correct
Solution: For a small temperature difference, $-\dfrac{dQ}{dt} \propto (T - T_0)$. It is an approximation to Stefan's law valid when $T - T_0$ is small.
Q9 — Thermal Radiation · medium · theory
Newton's law of cooling is a good approximation only when:
A. The surroundings are at absolute zero
B. The body is a perfect blackbody
C. The body is at a very high temperature
D. The temperature difference between the body and its surroundings is small ✓ Correct
Solution: Expanding Stefan's $T^4 - T_0^4$ law for small $T - T_0$ gives a linear dependence. For large differences the fourth-power law must be used instead.
Q10 — Thermal Radiation · hard · numerical
A body cools from $60^\circ\text{C}$ to $50^\circ\text{C}$ in $10\text{ minutes}$ in surroundings at $25^\circ\text{C}$. The time it takes to cool from $50^\circ\text{C}$ to $40^\circ\text{C}$ is:
A. $15\text{ minutes}$ ✓ Correct
B. $10\text{ minutes}$
C. $20\text{ minutes}$
D. $12\text{ minutes}$
Solution: By Newton's law, $\dfrac{\Delta T}{t} = k(T_{avg} - T_0)$. First stage: $\dfrac{10}{10} = k(55 - 25) = 30k \Rightarrow k = \dfrac{1}{30}$. Second stage: $\dfrac{10}{t} = k(45 - 25) = \dfrac{20}{30} \Rightarrow t = 15\text{ min}$.
Q11 — Thermal Radiation · easy · theory
A blackened surface placed in sunlight becomes hotter than a polished white surface because a black surface is:
A. A better absorber of radiation ✓ Correct
B. A better conductor of heat
C. A poorer emitter of radiation
D. A better reflector of radiation
Solution: A black surface absorbs nearly all the incident radiation instead of reflecting it, so it gains energy faster and its temperature rises more quickly.
Q12 — Thermal Radiation · easy · theory
The emissive power of a surface is defined as the:
A. Energy absorbed per unit area per unit time
B. Fraction of incident energy that is absorbed
C. Total energy radiated in the lifetime of the body
D. Energy radiated per unit area per unit time ✓ Correct
Solution: Emissive power is a radiant flux density, measured in $\text{W}/\text{m}^2$.
Q13 — Thermal Radiation · medium · theory
Ferry's blackbody is constructed as a double-walled hollow sphere with a small aperture because:
A. Radiation entering the aperture undergoes repeated reflections inside and is almost completely absorbed ✓ Correct
B. The sphere reflects all radiation that falls on the aperture
C. The small aperture prevents any radiation from entering
D. The double wall increases the reflectivity of the surface
Solution: Each internal reflection lets the blackened inner surface absorb a further fraction, so essentially nothing escapes back out of the tiny hole. The aperture therefore behaves as a near-perfect absorber.
Q14 — Thermal Radiation · medium · theory
Two bodies at the same temperature, one black and one polished, are placed in identical cool surroundings. The body that cools faster is:
A. Both cool at exactly the same rate
B. The polished body, since it reflects more
C. The black body, since it is the better emitter ✓ Correct
D. Neither, since both are at the same temperature
Solution: By Kirchhoff's law the better absorber is the better emitter. The black surface has the higher emissivity and so radiates its heat away more rapidly.
Q15 — Thermal Radiation · medium · theory
In the spectrum of a blackbody at a given temperature, the emitted energy:
A. Is concentrated entirely at a single wavelength
B. Is the same at every wavelength
C. Is distributed over all wavelengths with a distinct maximum at one wavelength ✓ Correct
D. Increases steadily with wavelength without limit
Solution: The blackbody spectrum is continuous, rising from zero, peaking at $\lambda_m$ and falling away again at long wavelengths.
Q16 — Thermal Radiation · easy · theory
Thermal radiation emitted by a hot body consists predominantly of:
A. Streams of electrons
B. Infrared electromagnetic waves ✓ Correct
C. Ultraviolet electromagnetic waves
D. Longitudinal sound waves
Solution: For bodies at ordinary and moderately high temperatures, the emission peak lies in the infrared region of the electromagnetic spectrum.
Q17 — Thermal Radiation · medium · theory
The greenhouse effect in the Earth's atmosphere arises because atmospheric gases:
A. Reflect all incoming solar radiation back to space
B. Conduct heat downward from the upper atmosphere
C. Transmit incoming short-wavelength sunlight but absorb outgoing long-wavelength infrared radiation ✓ Correct
D. Absorb incoming sunlight but transmit outgoing infrared radiation
Solution: Visible sunlight passes through and warms the surface, which then re-radiates in the infrared. Gases such as $\text{CO}_2$ absorb that infrared and re-emit part of it downward, warming the surface further.
Q18 — Thermal Radiation · hard · numerical
A body cools from $100^\circ\text{C}$ to $80^\circ\text{C}$ in $10\text{ minutes}$ in surroundings at $20^\circ\text{C}$. The time it takes to cool from $80^\circ\text{C}$ to $60^\circ\text{C}$ is:
A. $14\text{ minutes}$ ✓ Correct
B. $12\text{ minutes}$
C. $10\text{ minutes}$
D. $20\text{ minutes}$
Solution: By Newton's law, $\dfrac{\Delta T}{t} = k(T_{avg} - T_0)$. First stage: $\dfrac{20}{10} = k(90 - 20) = 70k \Rightarrow k = \dfrac{1}{35}$. Second stage: $\dfrac{20}{t} = \dfrac{50}{35} \Rightarrow t = 14\text{ min}$.
Q19 — Thermal Radiation · hard · numerical
A body of emissivity $0.6$ and surface area $0.1\text{ m}^2$ is at $500\text{ K}$. The power it radiates is ($\sigma = 5.67 \times 10^{-8}$):
A. $213\text{ W}$ ✓ Correct
B. $128\text{ W}$
C. $354\text{ W}$
D. $2126\text{ W}$
Solution: $P = e\sigma AT^4 = 0.6 \times 5.67 \times 10^{-8} \times 0.1 \times (500)^4 = 0.6 \times 5.67 \times 10^{-8} \times 0.1 \times 6.25 \times 10^{10} \approx 213\text{ W}$.
Q20 — Thermal Radiation · medium · numerical
A perfect blackbody at $400\text{ K}$ radiates energy at the rate of ($\sigma = 5.67 \times 10^{-8}$):
A. $227\text{ W/m}^2$
B. $1452\text{ W/m}^2$ ✓ Correct
C. $2268\text{ W/m}^2$
D. $459\text{ W/m}^2$
Solution: $E = \sigma T^4 = 5.67 \times 10^{-8} \times (400)^4 = 5.67 \times 10^{-8} \times 2.56 \times 10^{10} \approx 1452\text{ W/m}^2$.
Q21 — Thermal Radiation · hard · numerical
A blackbody of surface area $1\text{ m}^2$ at $400\text{ K}$ is placed in surroundings at $300\text{ K}$. The net rate of loss of radiant energy is ($\sigma = 5.67 \times 10^{-8}$):
A. $459\text{ W}$
B. $992\text{ W}$ ✓ Correct
C. $1452\text{ W}$
D. $1911\text{ W}$
Solution: $P_{net} = \sigma A(T^4 - T_0^4) = 5.67 \times 10^{-8}(2.56 \times 10^{10} - 8.1 \times 10^9) = 5.67 \times 10^{-8} \times 1.75 \times 10^{10} \approx 992\text{ W}$.
Q22 — Thermal Radiation · hard · numerical
A body cools from $50^\circ\text{C}$ to $40^\circ\text{C}$ in $5\text{ minutes}$ in surroundings at $20^\circ\text{C}$. The time it takes to cool from $40^\circ\text{C}$ to $30^\circ\text{C}$ is approximately:
A. $12.5\text{ minutes}$
B. $10.0\text{ minutes}$
C. $8.3\text{ minutes}$ ✓ Correct
D. $5.0\text{ minutes}$
Solution: By Newton's law, $\dfrac{\Delta T}{t} = k(T_{avg} - T_0)$. First stage: $\dfrac{10}{5} = k(45 - 20) = 25k \Rightarrow k = 0.08$. Second stage: $\dfrac{10}{t} = 0.08(35 - 20) = 1.2 \Rightarrow t \approx 8.3\text{ min}$.
Q23 — Thermal Radiation · hard · numerical
A body cools from $80^\circ\text{C}$ to $70^\circ\text{C}$ in $4\text{ minutes}$ in surroundings at $30^\circ\text{C}$. The time it takes to cool from $70^\circ\text{C}$ to $60^\circ\text{C}$ is approximately:
A. $5.1\text{ minutes}$ ✓ Correct
B. $4.0\text{ minutes}$
C. $8.0\text{ minutes}$
D. $6.5\text{ minutes}$
Solution: First stage: $\dfrac{10}{4} = k(75 - 30) = 45k \Rightarrow k = \dfrac{1}{18}$. Second stage: $\dfrac{10}{t} = \dfrac{35}{18} \Rightarrow t \approx 5.1\text{ min}$.
Q24 — Thermal Radiation · hard · numerical
A body of emissivity $0.8$ and surface area $0.5\text{ m}^2$ is at $300\text{ K}$. The power it radiates is ($\sigma = 5.67 \times 10^{-8}$):
A. $367\text{ W}$
B. $184\text{ W}$ ✓ Correct
C. $459\text{ W}$
D. $230\text{ W}$
Solution: $P = e\sigma AT^4 = 0.8 \times 5.67 \times 10^{-8} \times 0.5 \times 8.1 \times 10^9 \approx 184\text{ W}$.
Q25 — Thermal Radiation · medium · numerical
A perfect blackbody at $600\text{ K}$ radiates energy at the rate of ($\sigma = 5.67 \times 10^{-8}$):
A. $7348\text{ W/m}^2$ ✓ Correct
B. $3544\text{ W/m}^2$
C. $1452\text{ W/m}^2$
D. $459\text{ W/m}^2$
Solution: $E = \sigma T^4 = 5.67 \times 10^{-8} \times (600)^4 = 5.67 \times 10^{-8} \times 1.296 \times 10^{11} \approx 7348\text{ W/m}^2$.
Q26 — Thermal Radiation · hard · numerical
A blackbody of area $2\text{ m}^2$ at $500\text{ K}$ sits in surroundings at $300\text{ K}$. The net rate of radiation loss is ($\sigma = 5.67 \times 10^{-8}$):
A. $7088\text{ W}$
B. $6169\text{ W}$ ✓ Correct
C. $3084\text{ W}$
D. $918\text{ W}$
Solution: $P = \sigma A(T^4 - T_0^4) = 5.67 \times 10^{-8} \times 2 \times (6.25 \times 10^{10} - 8.1 \times 10^9) \approx 6169\text{ W}$.
Q27 — Thermal Radiation · easy · numerical
Two blackbodies are at $200\text{ K}$ and $400\text{ K}$. The ratio of their emissive powers is:
A. $1 : 2$
B. $1 : 8$
C. $1 : 4$
D. $1 : 16$ ✓ Correct
Solution: $E \propto T^4$, so the ratio is $\left(\dfrac{200}{400}\right)^4 = \dfrac{1}{16}$.
Q28 — Thermal Radiation · medium · numerical
A perfect blackbody at $727^\circ\text{C}$ radiates energy at the rate of ($\sigma = 5.67 \times 10^{-8}$):
A. $5670\text{ W/m}^2$
B. $56700\text{ W/m}^2$ ✓ Correct
C. $23224\text{ W/m}^2$
D. $7348\text{ W/m}^2$
Solution: $T = 727 + 273 = 1000\text{ K}$, so $E = 5.67 \times 10^{-8} \times 10^{12} = 5.67 \times 10^4\text{ W/m}^2$.
Q29 — Thermal Radiation · easy · numerical
The absolute temperature of a blackbody rises from $300\text{ K}$ to $600\text{ K}$. Its radiated power increases by a factor of:
A. $8$
B. $2$
C. $16$ ✓ Correct
D. $4$
Solution: $E \propto T^4$ and the temperature doubles, so the power rises by $2^4 = 16$ times.
Q30 — Thermal Radiation · hard · numerical
A blackbody of surface area $0.02\text{ m}^2$ is maintained at $800\text{ K}$. The power it radiates is ($\sigma = 5.67 \times 10^{-8}$):
A. $23224\text{ W}$
B. $928\text{ W}$
C. $464\text{ W}$ ✓ Correct
D. $232\text{ W}$
Solution: $P = \sigma AT^4 = 5.67 \times 10^{-8} \times 0.02 \times 4.096 \times 10^{11} \approx 464\text{ W}$.