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Question: A student is trying to decide what to wear. His bedroom is at \({20^0}C\) His skin temperature is \(...

A student is trying to decide what to wear. His bedroom is at 200C{20^0}C His skin temperature is 350C{35^0}C The area of his exposed skin is 1.50m21.50\,{m^2} People all over the world have skin that is dark in the infrared, with emissivity about 0.9000.900 Find the net energy transfer from his body by radiation in 10.0min10.0\min .

Explanation

Solution

In order to solve this question, we will first calculate the power transferred from the skin of the student using the general formula of power emitted in case of black body radiation and then we will calculate the energy emitted by the body of student for given period of time.

Formula used:
Power emitted by a body with a temperature difference from surface area is calculated as
P=σAe(T4T04)P = \sigma Ae({T^4} - {T_0}^4)
where, σ=5.67×108Wm2K4\sigma = 5.67 \times {10^{ - 8}}W{m^{ - 2}}{K^{ - 4}} known as Stefan constant, AA is area which is given in question as A=1.50m2A = 1.50{m^2}, e=0.900e = 0.900 known as emissivity constant, TT is the temperature of the body given in question as T=350CT = {35^0}C and T0=200C{T_0} = {20^0}C the temperature of surrounding of the body.

Complete answer:
Now to calculate power emitted by the body of student, using the formula P=σAe(T4T04)P = \sigma Ae({T^4} - {T_0}^4) on putting the value we get,
P=5.67×108×1.50×0.900(T4T04)P = 5.67 \times {10^{ - 8}} \times 1.50 \times 0.900({T^4} - {T_0}^4)
converting temperature from Celsius to kelvin we have,
T=35+273=308KT = 35 + 273 = 308K
T0=20+273=293K\Rightarrow {T_0} = 20 + 273 = 293K
Now,
P=5.67×108×1.50×0.900((308)4(293)4)P = 5.67 \times {10^{ - 8}} \times 1.50 \times 0.900({(308)^4} - {(293)^4})
P=125W\Rightarrow P = 125\,W
Now, in order to calculate the energy emitted by body in time of t=10.0min=600sect = 10.0\min = 600\sec we know that,
E=P×tE = P \times t
so we have,
E=125×600JE = 125 \times 600J
E=72KJ\therefore E = 72\,KJ

Hence, the net energy transferred in period of ten minutes by the body of student is E=72KJE = 72\,KJ.

Note: It should be remembered that, the basic conversion rule of Celsius scale to kelvin scale is given as K=0C+273K{ = ^0}C + 273 and Kilo-Joule is a unit of energy written as KJ and its value is 1KJ=1000J1KJ = 1000J also, In the infrared region of light a normal person shines much brighter than a hundred-watt light bulb.