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(Solved): The emissivity of the human skin is \( 97.0 \) percent. Use \( 35.0^{\circ} \mathrm{C} \) for the ...



The emissivity of the human skin is \( 97.0 \) percent. Use \( 35.0^{\circ} \mathrm{C} \) for the skin temperature and approx

The emissivity of the human skin is \( 97.0 \) percent. Use \( 35.0^{\circ} \mathrm{C} \) for the skin temperature and approximate the human body by a rectangular block with a height of \( 1.55 \mathrm{~m} \), a width of \( 30.0 \mathrm{~cm} \) and a length of \( 34.0 \mathrm{~cm} \). Calculate the Dower emitted by the human body. Tries \( 0 / 12 \) What is the wavelength of the peak in the spectral distribution for this temperature? Tries \( 0 / 12 \) Fortunately our environment radiates too. The human body absorbs this radiation with an absorptance of \( 97.0 \) percent, so we don't lose our internal energy so quickly. How much power do we absorb when we are in a room where the temperature is \( 20.0^{\circ} \mathrm{C} \) ? Tries \( 0 / 12 \) How much enerqy does our body lose in one second? Tries \( 0 / 12 \)


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PartA) Given data: ?=97.0 is the emissivity of the human skin, T=273+35=308 K is the temperature of the human body in Kelvin. Height of rectangle=1.55
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