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 $T_{o}$. Take the room temperature to be $T_{o} = 300K$. For $T_{o} = 300K$, the value of $\sigma T_{o}^{4} = 460 Wm^{-2}$ (where $\sigma$ is Stefan-Boltzmann constant). What of the following options is/are correct?
a) If the surrounding temperature reduces by a small amount $\Delta T_{o} \lt \lt T_{o}$, then to maintain the same body temperature the same (living) human being needs to radiate $\Delta W = 4 \sigma T_{o}^{3} \Delta T_{o}$ more energy per unit time
b) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths.
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) The amount of energy radiated by the body in $1$ second is close to $60$ joules.

Answer 1

Hint: When human body is treated as a black body then we will use Stephan’s law (It states that radiated power density is directly proportional to fourth power of temperature) and Wein’s displacement law (It states that wavelength is directly proportional to temperature) to solve the problem.

Complete answer:
Given Area = $1 m^{2}$
​​$T_{b} = T_{o} + 10$ ($T_{b}$ is temperature of body)
Also, $\sigma T_{o}^{4} = 460 Wm^{-2}$
a) $W = \sigma A (T_{b}^{4} - T_{o}^{4}) $
$W = \sigma A (T_{b}^{4} – (T_{o} – \Delta T_{o})^{4}) $
Using binomial approximation, we get,
$W^{‘} = \sigma A (T_{b}^{4} – (T_{o}^{4} –4 \sigma T_{o}^{3} \Delta T_{o}) $ (other terms will be neglected)
Put $W = \sigma A (T_{b}^{4} - T_{o}^{4}) $ in the above equation.
Hence, $ W^{‘} = W + \sigma (T_{b}^{4} - T_{o}^{4}) $
b) We know that $\lambda T = constant$
Hence if the temperature of a body is increased the wavelength at the peak point will shift to a lower wavelength.
c) $W = \sigma A (T_{b}^{4} - T_{o}^{4}) $
If the surface area of the body will reduce, then energy will be minimum. Reducing the exposed surface area of the body allows humans to maintain the same body temperature while reducing the energy lost by radiation.
d) $W = \sigma A (T_{b}^{4} - T_{o}^{4}) $
Since $\sigma T_{o}^{4} = 460 Wm^{-2}$ and $T_{o} = 300K$
Put $\sigma = \dfrac{460}{300^{4}}$, $T_{b} = 310K$ and $T_{o} = 300K$
$W = \dfrac{460}{300^{4}}(310^{4} - 300^{4}) $
$W = 64.46 J s^{-1}$ which is close to $60 J s^{-1}$

So, the correct answer is “Option a,c and d”.

Note: We treated the human body as a perfect black body, if the body is not a perfect black body then emissivity term comes into play into Stephan’s law formula. Emissivity e is an improvement for an estimated black body radiator, where $e = 1 – R$, R is the portion of the light reflected by the black body.