Question #57318

person with external body temperature 35degree celcius in room temp 25degree celsius emisivity 0.5 surface area 2.0m2 calculate radiant power
1

Expert's answer

2016-01-19T08:28:41-0500

Answer on Question 57318, Physics, Mechanics, Relativity

Question:

A person with external body temperature 35C35{}^{\circ}\mathrm{C} is present in a room at temperature 25C25{}^{\circ}\mathrm{C}. Assuming the emissivity of the body of the person to be 0.5 and surface area of the body of the person as 2.0m22.0m^2, calculate the radiant power of the person.

Solution:

The person radiates energy at a rate:


P=QΔt=εσA(T14T24),T1>T2P = \frac{Q}{\Delta t} = \varepsilon \sigma A (T_1^4 - T_2^4), T_1 > T_2


here, PP is the radiant power of the person, ε=0.5\varepsilon = 0.5 emissivity of the body of the person, σ=5.672108Jsm2K4\sigma = 5.672 \cdot 10^{-8} \frac{J}{s \cdot m^2 \cdot K^4} is the Stefan-Boltzmann constant, A=2.0m2A = 2.0m^2 is the surface area of the body of the person and T1T_1 is the temperature of the person, and T2T_2 is the temperature of the surroundings.

Then, the radiant power of the person will be:


P=εσA(T14T24)==0.55.672108Jsm2K42.0m2((35+273.15K)4(25+273.15K)4)=63.2W.\begin{array}{l} P = \varepsilon \sigma A (T_1^4 - T_2^4) = \\ = 0.5 \cdot 5.672 \cdot 10^{-8} \frac{J}{s \cdot m^2 \cdot K^4} \cdot 2.0m^2 \\ \cdot \left((35 + 273.15K)^4 - (25 + 273.15K)^4\right) = 63.2W. \end{array}


Answer:


P=63.2W.P = 63.2W.


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