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Chapter 16: Problem 38

A typical human body has surface area \(1.4 \mathrm{m}^{2}\) and skin temperature \(33^{\circ} \mathrm{C} .\) If the body's emissivity is about \(1,\) what's the net radiation from the body when the ambient temperature is \(18^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The net radiation from the body when the ambient temperature is \(18^{\circ} \mathrm{C}\) is \(102.2 \, W\).

Step by step solution

01

Convert Temperatures to Kelvin

Since the Stefan-Boltzmann law requires temperatures to be measured in Kelvin, you should first convert the temperatures from degrees Celsius to Kelvin. Add 273.15 to each of the Celsius temperature values. This gives skin temperature \(T_s = 33 + 273.15 = 306.15 \, K\) and ambient temperature \(T_a = 18 + 273.15 = 291.15 \, K\).
02

Apply Stefan-Boltzmann Law

With the temperatures expressed in Kelvin, you can use the Stefan-Boltzmann law to calculate the radiant power per square meter for the human body and the surroundings separately. This law is formulated as \(j = \sigma T^4\), where \(j\) is the radiative power per unit area, \(\sigma = 5.67 \times 10^{-8} \, W/m^2K^4\) is the Stefan-Boltzmann constant, and \(T\) is the temperature in Kelvin. For the body: \(j_s = 1 \times 5.67 \times 10^{-8} \times (306.15)^4 = 474.5 \, W/m^2\), and for the surroundings: \(j_a = 1 \times 5.67 \times 10^{-8} \times (291.15)^4 = 401.5 \, W/m^2\)
03

Calculate Net Radiation

To obtain the net radiation, subtract the radiation from the surroundings (\(j_a\)) from that of the body (\(j_s\)), and then multiply by the surface area of the body. That is, \(net \,radiation = (j_s - j_a)A = (474.5 - 401.5) \times 1.4 = 102.2 \, W\)

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