Find the net rate of radiation by a pot at 46.1 ∘C that has been placed...
Find the net rate of radiation by a pot at 46.1 ∘C that has been placed in a −20.7 ∘C freezer. The pot's surface area is 0.131 m2and its emissivity is 0.581.
net rate of radiation:
Homework Answers
Stefan-Boltzmann constant = 
= 5.67 x 10-8 W/(m2.K4)
Surface area of the pot = A = 0.131 m2
Emissivity of the pot = e = 0.581
Temperature of the pot = T1 = 46.1 oC = 46.1 + 273 K = 319.1 K
Temperature of the freezer = T2 = -20.7 oC = -20.7 + 273 K = 252.3 K
Net rate of radiation from the pot = Q
Q = Ae(T14
- T24)
Q = (0.131)(0.581)(5.67x10-8)(319.14 - 252.34)
Q = 27.26 W
Net rate of radiation from the pot = 27.26 W
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