Question

A human body has a surface area of approximately $1\, m^2$. The normal body temperature is $10\, K$ above the surrounding room temperature $

• A. The amount of energy radiated by the body in $1$ second is close to $60$ Joules
• B. If the surrounding temperature reduces by a small amount $\Delta\,<
• C. Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation
• D. If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

Answer 1

The Correct Option is C

Rate of energy loss due to radiation
$\frac{ dQ }{ dt }=\sigma AT ^{4}$
Rate of energy absorbed by surrounding
$\frac{ dQ }{ dt }=\sigma e AT _{0}^{4}$
Net heat loss by radiation $\frac{ dQ }{ dt }=\sigma eA\left( T ^{4}- T _{0}^{4}\right)$
If exposed area A decreases. Rate of heat loss also decreases.
(B)

If body temperature rises, spectrum of electromagnetic radiation shifts to smaller wavelength.
(C) $\frac{ dQ }{ dt } = \sigma eA(T^4 - T_0^4)$
$= \sigma eA\left[(T_0 + \Delta T)^4-T_0^4\right]$
$= \sigma eAT_{0}^{4}\left[\left(1+\frac{\Delta T }{ T _{0}}\right)^{4}-1\right]$
$=\sigma eA T _{0}^{4}\left[1+\frac{4 \Delta T }{ T _{0}}-1\right]$
$=\sigma eAT_{0}^{4}\left(\frac{4 \Delta T }{ T _{0}}\right)$
$=1 \times 460\left(\frac{4 \times 10}{300}\right)=\frac{184}{3}=61.3 J$