Question

If the absolute temperature of a black body is doubled, then the percentage increase in the rate of loss of heat by radiation is :

• A. 15%
• B. 16%
• C. 1600%
• D. 1500%

Answer 1

Answer: (D)

1500%

The rate of loss of heat by radiation is given by the Stefen-boltzmann law as:

$Q=\sigma T^{4}A$ where $\sigma$ is the stefen-boltzmann constant, $A$ is the area of the radiating body.

So $Q\propto T^4$.

If $T$ is doubled, $Q$ becomes $2^4=16$ times.

So, percentage increase in rate of heat loss by radiation is $(16Q-Q)\times 100/Q=1500\%$