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 = $$\sigma$$eAT⁴,

where $$\sigma$$ is Stefan's constant. If this body is kept at a surrounding temperature T₀ 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 $$\Delta$$T₀ (<< T₀) 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 $$\Delta$$T₀ 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.