Question

The human body has a surface area of approximately $1{{m}^{2}}$. The normal body temperature is $10K$ above the surrounding room temperature ${{T}_{0}}$. Take the room temperature to be ${{T}_{0}}=300K$. For ${{T}_{0}}=300K$ and $\sigma T_{0}^{4}=460W{{m}^{-2}}$ (where $\sigma $ is the Stefan-Boltzmann constant). Which of the following is/are correct?
(A). 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
(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 (eg. 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 sec is close to 60 Joules

Answer 1

Hint: The Stefan-Boltzmann law can be applied to the heat emitted by a human body. According to this law, the heat emitted depends on the area and temperature. The Wein’s displacement law gives the relation between temperature and wavelength of the heat emitted.
Formulas used:
$\Delta W=\sigma A(T_{b}^{4}-T_{0}^{4})$
$\lambda T=k$

Complete answer:
Given that, area of a human body is $1{{m}^{2}}$, The temperature of the body is given by-
$\begin{align}
 & {{T}_{b}}={{T}_{0}}+10 \\
 & \Rightarrow {{T}_{b}}=300+10=310K \\
\end{align}$
$\sigma T_{0}^{4}=460W{{m}^{-2}}$
Given values are substituted in the above equation to get,
$\begin{align}
 & \sigma T_{0}^{4}=460W{{m}^{-2}} \\
 & \Rightarrow \sigma {{(300)}^{4}}=460 \\
\end{align}$
$\therefore \sigma =5.67\times {{10}^{-8}}W\,{{m}^{-2}}\,{{K}^{-4}}$
According to the Stefan-Boltzmann law, the total energy radiated per unit surface area from a black body is directly proportional to the fourth power of temperature. Therefore, the energy radiated from the human body. Therefore, the difference of heat radiated is given as-
$\Delta W=\sigma A(T_{b}^{4}-T_{0}^{4})$ - (1)
Temperature of the surroundings changes by a very small amount $\Delta {{T}_{0}}<<{{T}_{0}}$, we substitute it in the above equation to get,
$\Delta W=\sigma (T_{b}^{4}-{{({{T}_{0}}-\Delta {{T}_{0}})}^{4}})$
Using Binomial approximation in the above equation we get,
$\Delta W=\sigma (T_{b}^{4}-(T_{0}^{4}-4T_{0}^{3}\Delta {{T}_{0}}))$ (other terms will be neglected)
$\begin{align}
 & \Delta W=\sigma (T_{b}^{4}-(T_{0}^{4}-4T_{0}^{3}\Delta {{T}_{0}})) \\
 & \Rightarrow \Delta W=\sigma (T_{b}^{4}-(T_{0}^{4}-4T_{0}^{3}\Delta {{T}_{0}})) \\
\end{align}$
$\therefore \Delta W=\sigma 4T_{0}^{3}\Delta {{T}_{0}}$ (other terms will be negligible)
Therefore, the heat emitted per unit area per unit time will be $4\sigma T_{0}^{3}\Delta {{T}_{0}}$ more than the original amount of heat emitted.
According to the Wein’s law, the wavelength of the radiation emitted is inversely proportional to the temperature, therefore,
$\lambda T=k$
Here, $\lambda $ is the wavelength
$T$ is the temperature
$k$ is constant
Therefore, if the temperature rises the peak of radiation will shift to lower wavelengths rather than longer wavelengths.
From eq (1), the heat emitted depends on the area therefore; if the area decreases, heat radiated will also decrease and hence the same body temperature can be maintained by reducing the radiation.
We know from eq (1),
$\Delta W=\sigma A(T_{b}^{4}-T_{0}^{4})$
Given, $\sigma T_{0}^{4}=460W{{m}^{-2}}$ and $\sigma =5.67\times {{10}^{-8}}W\,{{m}^{-2}}\,{{K}^{-4}}$
Given values are substituted in the above equation to get,
$\Delta W\approx 64.46J\,{{s}^{-1}}$
Therefore, the value of heat emitted per second is close to 60 Joules.
Therefore, the heat emitted per unit area per unit time will be $4\sigma T_{0}^{3}\Delta {{T}_{0}}$ more than the original amount of heat emitted, if the temperature rises the peak of radiation will shift to lower wavelengths rather than longer wavelengths and the value of heat emitted per second is close to 60 Joules.

So, the correct answer is “Option A, C and D”.

Note: Stefan-Boltzmann constant is the constant of proportionality in the Stefan-Boltzmann law. An ideal black body is the one which radiates heat over all wavelengths. The Stefan Boltzmann law describes the heat radiation given out by hot bodies and is not limited to this case only.