Question

a person with external body temperature 3 degree Celsius is present in a room at temperature 25 degree Celsius . assuming the emissivity of the body of the person to be 0.5 and surface area of the body of the person as 2.0 m square , calculate the radiant power of the person

Answer 1

Answer on Question 60518, Physics, Mechanics, Relativity

Question:

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

Solution:

We can find the radiant power of the person from the Stefan-Boltzmann Law:

$$P = \frac {Q}{\Delta t} = \varepsilon \sigma A (T _ {1} ^ {4} - T _ {2} ^ {4}), T _ {1} > T _ {2}$$

here, $P$ is the radiant power of the person, $\varepsilon = 0.5$ is the emissivity of the body of the person, $\sigma = 5.672 \cdot 10^{-8} \frac{J}{s \cdot m^2 \cdot K^4}$ is the Stefan-Boltzmann constant, $A = 2.0m^2$ is the surface area of the body of the person, $T_1$ is the temperature of the person, and $T_2$ is the temperature of the surroundings.

Then, from this formula we can calculate the radiant power of the person:

$$\begin{array}{l} P = \varepsilon \sigma A (T _ {1} ^ {4} - T _ {2} ^ {4}) = \\ = 0.5 \cdot 5.672 \cdot 10^{-8} \frac{J}{s \cdot m^2 \cdot K^4} \cdot 2.0m^2 \\ \cdot \left((35 + 273.15K)^4 - (25 + 273.15K)^4\right) = 63.2W. \end{array}$$

Answer:

$$P = 63.2W.$$

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