The net rate of radiant heat flow from an object is given by Delta Q / Delta t =
ID: 1956271 • Letter: T
Question
The net rate of radiant heat flow from an object is given by Delta Q / Delta t = e A (x14 - t24) where A is the area of the object, T, its temperature, e its emissivity (value depends on surface), and T2 the temperature of its surroundings. = 5.67 times 10 -8 WM -2K -4. A person has a skin temperature of 33 Degree C and is in a room when the walls have a temperature of 29 Degree C. If the emissivity for skin is 0.70 (e-0.70) and if the body surface area is 1.5M2 , then calculate Delta Q / Delta t net, the net radiative heat loss in watts.Explanation / Answer
eA(T14 - T24) = (0.70)(5.67*10-8)(1.5)[(273+33)4 - (273+29)4]
= 26.76 W
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