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# If the total surface area of a human body is 1.2$m^2$ and the surface temperature is 30$^{\circ}$C, find the net rate of radiation from the body if the surrounding is at a temperature of 20$^{\circ}$C. (Take emissivity of human body = 1)A. 574 WB. 72 WC. 800 WD. 60 W

Last updated date: 13th Jun 2024
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Hint: As the human body is acting like a black body and is emitting radiations, the answer can be obtained by using Stefan-Boltzmann law for black body radiation. If the body is in an enclosure at some temperature with respect to the body, the difference of the two temperatures (power 4) has to be taken.
Formula used:
Energy radiated by a surface of emissivity $\epsilon$, area A in time t,
$E = \epsilon \sigma (T^4 - T_0^4) At$

First, let us convert the temperatures in Kelvin by adding 273 to Celsius temperature, we have
T = 273 + 30 = 303 K and
T_0 = 20 + 273 = 293 K.
Also, we are given,
$A = 1.2m^2$
$\epsilon = 1$
and we know
$\sigma = 5.67 \times 10^{-8} J m^{-2} s^{-1} K^{-4}$.

The rate of energy radiation will be given as:
$E = \dfrac{ E'}{t}$
where E' has been used here to denote the energy emitted in time t (also written as E in the formula used section).
This will give us the rate as:
$E = \epsilon \sigma (T^4 - T_0^4) A$.
We substitute our values here, we get:
$E = 5.67 \times 10^{-8} \times [(303)^4 - (293)^4] \times 1.2 = 71.986 W$.

So, the correct answer is “Option B”.