Question

If the temperature of a hot body is raised by $0.5\%$, then the heat energy radiated would increase by :

• A. 0.5%
• B. 1.0%
• C. 1.5%
• D. 2.0%

Answer 1

The correct option is D 2.0%

The rate of heat energy radiated by black body is given as:

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

So $Q\propto T^4=k{T}^4$, where $k$ is the propotionality constant that we have assumed here.

Taking log of above equation, $logQ=log\left(k{T}^4\right)=logk+4logT$

Taking differential of above equation, $\frac{dQ}{Q}=0+\frac{4dT}{T}$ (because logk is constant).

Here $dQ$ and $dT$ indicate very small change in the $Q$ and $T$ respectively.

Multiplying by $100$,

$\frac{dQ}{Q}\times 100=4\frac{dT}{T}\times 100$

$\Rightarrow$ Percentage change in Heat rate $=4\times$ percentage change in temperature

So here, percentage change in rate of heat energy radiated $=4\times 0.5=2\%$