Answer
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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,
$E = \epsilon \sigma (T^4 - T_0^4) At$
Complete answer:
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”.
Additional Information:
Black body definition is a perfect absorber and emitter of radiation. The colour of the black body doesn't have to be black. Like in our case, we assumed the human body to be a black body. It happens so that if no light reflects from a body and all the light falling on it gets absorbed by it, we perceive the colour of the body to black. It is for an ideal situation.
Note:
There are different expressions for Stefan-Boltzmann law for blackbody radiation. So, one must really go through the problem statement carefully to pick out the right expression for use. We are asked to find the rate of radiation here. The definition of the law contains energy emitted per second per unit area by a black body. By this definition, there is no factor of A or t in the expression for energy. But, when we talk of total energy radiated in time t, by a body of surface area A, then we use the expression of the form that we have under the 'formula used' section.
Formula used:
Energy radiated by a surface of emissivity $\epsilon$, area A in time t,
$E = \epsilon \sigma (T^4 - T_0^4) At$
Complete answer:
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”.
Additional Information:
Black body definition is a perfect absorber and emitter of radiation. The colour of the black body doesn't have to be black. Like in our case, we assumed the human body to be a black body. It happens so that if no light reflects from a body and all the light falling on it gets absorbed by it, we perceive the colour of the body to black. It is for an ideal situation.
Note:
There are different expressions for Stefan-Boltzmann law for blackbody radiation. So, one must really go through the problem statement carefully to pick out the right expression for use. We are asked to find the rate of radiation here. The definition of the law contains energy emitted per second per unit area by a black body. By this definition, there is no factor of A or t in the expression for energy. But, when we talk of total energy radiated in time t, by a body of surface area A, then we use the expression of the form that we have under the 'formula used' section.
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