Question

In this problem you will consider the balance of thermal energy radiated and absorbed by a person in a room. The rate of heat transfer from radiation is: ΔQΔt=eσA(T42−T41)=eσAT42−eσAT41. This equation has two terms which represent the rate of absorption from the room (a gain) and the rate of radiation into the room (a loss). In this problem, we will consider these two terms separately. Assume that the area of a human body may be considered to be that of the sides of a cylinder of length L=2.0m and circumference C=0.8m. For the Stefan-Boltzmann constant use σ=5.67×10−8W/m2⋅K4. The rate of heat loss to the room can be calculated using only the first term of the equation ( ΔQΔt=eσAT4). If the surface temperature of the body is taken to be Tbody=30∘C , what is the rate of heat loss by the body? Take the emissivity to be 0.6 and assume that we are only considering losses to the room (not gain from the room). Express the power radiated into the room by the body numerically, rounded to the nearest 10 W.

Answer 1