Question 410: Properties of Matter

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 $${T_0}.$$ Take the room temperature to be $${T_0} = 300K.$$ For $${T_0} = 300\,K,$$ the value of $$\sigma T_0^4 = 460\,W{m^{ - 2}}$$ (where $$\sigma $$ is the Stefan-Boltzmann constant).
Which of the following options is / are correct?

Accepted Answer

The rate of radiation energy emitted by a body of surface area A, emissivity e, and kept at an absolute temperature T is given by Stefan's law
dQ
e
/dt = $$\sigma$$eAT
4
,
where $$\sigma$$ is Stefan's constant. If this body is kept at a surrounding temperature T
0
then rate of radiation energy absorbed by the body is
dQ
a
/dt = $$\sigma$$eAT$$_0^4$$.
Thus, the net rate of energy loss by the body is
$$d{Q_i}/dt = d{Q_e}/dt - d{Q_a}/dt = \sigma eA({T^4} - T_0^4)$$.
If surrounding temperature is reduced by $$\Delta$$T
0
(<< T
0
) then net rate of energy lost by the body becomes
$${{d{Q_f}} \over {dt}} = \sigma eA\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$$
$$ = \sigma eA\left[ {{T^4} - T_0^4{{\left( {1 - {{\Delta {T_0}} \over {{T_0}}}} \right)}^4}} \right]$$
$$ \approx \sigma eA\left[ {{T^4} - T_0^4\left( {1 - 4{{\Delta {T_0}} \over {{T_0}}}} \right)} \right]$$
$$ = {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0}$$.
Thus, decrease in the surrounding temperature increases the rate of energy loss by $$4\sigma eAT_0^3\Delta {T_0}$$. To maintain the body temperature, the human being needs to increase the rate of energy generation (through chemical reactions in the body) by $$4\sigma eAT_0^3\Delta {T_0}$$ to balance the increase in rate of energy loss.
The human beings can also maintain the body temperature (without increasing the rate of energy generation) by reducing the surface area (e.g., by curling up) of their body. If surrounding temperature is reduced by $$\Delta$$T
0
and surface area of the body is reduced by $$\Delta$$A (<< A) then net rate of energy loss by the body becomes
$${{d{Q_f}} \over {dt}} = 4\sigma e(A - \Delta A)\left[ {{T^4} - {{({T_0} - \Delta {T_0})}^4}} \right]$$
$$ \approx {{d{Q_i}} \over {dt}} + 4\sigma eAT_0^3\Delta {T_0} - \sigma e\Delta A({T^4} - \Delta T_0^4)$$.
Thus, the rate of energy loss by radiation remains same $$(d{Q_f}/dt = d{Q_i}/dt)$$ if reduction in surface area is
$$\Delta A = {{4AT_0^3\Delta {T_0}} \over {{T^4} - T_0^4}} \approx A{{\Delta {T_0}} \over {\Delta T}}$$,
where, the body temperature is $$T = {T_0} + \Delta T$$.
We assume body to be black i.e., e = 1. The amount of energy emitted by the body in one second is
$$d{Q_e}/dt = \sigma eA{T^4} = \sigma eA{({T_0} + \Delta T)^4}$$
$$ \approx \sigma eAT_0^4(1 + 4\Delta T/{T_0})$$
$$ = (1)(1)(460)(1 + 4(10)/300) = 521$$ J.
The amount of energy lost by the body in one second is
$$d{Q_i}/dt = \sigma eA({T^4} - T_0^4) = \sigma eA{({T_0} + \Delta T)^4} - T_0^4)$$
$$ \approx eA\sigma T_0^4(4\Delta T/{T_0}) = 61$$ J.

Answer

The amount of energy radiated by the body in $$1$$ second is close to $$60$$ joules

Answer

If the surrounding temperature reduces by a small amount $$\Delta {T_0} <  < {T_0},$$ then to maintain the same body temperature the same (living) human being  needs to radiate  $$\Delta W = 4\sigma T_0^3\Delta {T_0}$$ more energy per unit time

Answer

Reducing the exposed surface area of the body (e.g. by curling up) allows human to maintain the same body temperature while reducing the energy lost by radiation

Answer

If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths

Properties of Matter

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