# ohys101(heat)

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a person had a skin temperature of 33degrees celcuis and was in a room where the walls were at 20degrees celcuis.if the emissivity is 1 and the body surface is 1.5meter squad,what is the rate of heat loss due to radiation? alpha=0.000000567W/meter squad k exponent 4

• ohys101(heat) -

Convert those temperatures to Kelvin. Call them T1(body) and T2(room walls)

The heat loss rate is

Q = (emissivity)*A*(alpha)(T1^4 - T2^4)

where A is the area (1.5 m^2 in this case)

Using this formula implies that the walls have an emissivity of 1 also. The body absorbs a fraction called the absorptivity, which equals the emissivity - assuming no wavelength dependence. That's one of Kirchhoff's radiation laws. There's a lot more to this subject than they are teaching you.

• ohys101(heat) -

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