Question

The rate of heat radiation from two patches of skin of equal area on a patient's chest differs by $2\%$. If one of the patches is at $300K$ and both the patches is assumed to be blackbody, the temperature of other patch will be

• A. $306\text{K}$
• B. $312\text{K}$
• C. $308.5\text{K}$
• D. $301.5\text{K}$

Answer 1

The correct option is D $301.5\text{K}$

According to Stefan-Boltzmann law,
Rate of heat radiation, $P=\sigma AϵT^4$
$\Rightarrow P=\sigma A{T}^4$
$[$ for blackbody, emissivity, $ϵ=1]$
So, $\Delta P=\sigma A4T^3\Delta T$
$\Rightarrow \frac{\Delta P}{P}=\frac{\sigma A4T^3\Delta T}{P}$
$\Rightarrow \frac{\Delta P}{P}=\frac{\sigma A4T^3\Delta T}{\sigma A{T}^4}$
$\Rightarrow \frac{\Delta P}{P}=\frac{4\Delta T}{T}$
$\Rightarrow \frac{2}{100}=\frac{4\Delta T}{300}$
$[$ given, rate of heat radiation differ by $2\%,T=300K]$
$\Rightarrow \Delta T=\frac{2\times 300}{100\times 4}=1.5\text{K}$
$\therefore$ Temperature of other patch will be $T+\Delta T=300+1.5=301.5\text{K}$