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Question

Physics Question on thermal properties of matter

A human body has a surface area of approximately 1m21\, m^2. The normal body temperature is 10K10\, K above the surrounding room temperature $

A

The amount of energy radiated by the body in 11 second is close to 6060 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

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

Explanation

Solution

Rate of energy loss due to radiation
dQdt=σAT4\frac{ dQ }{ dt }=\sigma AT ^{4}
Rate of energy absorbed by surrounding
dQdt=σeAT04\frac{ dQ }{ dt }=\sigma e AT _{0}^{4}
Net heat loss by radiation dQdt=σeA(T4T04)\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) dQdt=σeA(T4T04)\frac{ dQ }{ dt } = \sigma eA(T^4 - T_0^4)
=σeA[(T0+ΔT)4T04]= \sigma eA\left[(T_0 + \Delta T)^4-T_0^4\right]
=σeAT04[(1+ΔTT0)41]= \sigma eAT_{0}^{4}\left[\left(1+\frac{\Delta T }{ T _{0}}\right)^{4}-1\right]
=σeAT04[1+4ΔTT01]=\sigma eA T _{0}^{4}\left[1+\frac{4 \Delta T }{ T _{0}}-1\right]
=σeAT04(4ΔTT0)=\sigma eAT_{0}^{4}\left(\frac{4 \Delta T }{ T _{0}}\right)
=1×460(4×10300)=1843=61.3J=1 \times 460\left(\frac{4 \times 10}{300}\right)=\frac{184}{3}=61.3 J