Question:

A black body at a temperature of 125^\circC emits heat at the rate of 32 W. The rate of heat emitted by the body when the temperature of the body is increased by 398 K is:

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The power radiated by a black body increases with the fourth power of its temperature. When the temperature increases, the rate of heat emission increases significantly.
Updated On: Mar 12, 2025
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The Correct Option is C

Solution and Explanation

Step 1:
The rate of heat emission from a black body is proportional to the fourth power of its temperature: PT4 P \propto T^4
Step 2:
Let the initial temperature be T1=125C=125+273=398K T_1 = 125 ^\circ C = 125 + 273 = 398 \, K , and the final temperature be T2=398+398=796K T_2 = 398 + 398 = 796 \, K .
Step 3:
Using the Stefan-Boltzmann law: P2P1=(T2T1)4 \frac{P_2}{P_1} = \left( \frac{T_2}{T_1} \right)^4 P232=(796398)4 \frac{P_2}{32} = \left( \frac{796}{398} \right)^4 P232=24=16 \frac{P_2}{32} = 2^4 = 16 P2=32×16=512W m2 P_2 = 32 \times 16 = 512 \, \text{W m$^-2$}
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