Answer:
4440 W
Explanation:
k = Heat conduction coefficient = 0.04 W/(m·°C)
A = Area = 370 m²
l = Thickness = 10 cm
[tex]\Delta T[/tex] = Difference in temperature = 30°C
Rate of heat transfer is given by
[tex]Q=\dfrac{kA(T_2-T_1)}{l}\\\Rightarrow Q=\dfrac{0.04\times 370\times 30}{10\times 10^{-2}}\\\Rightarrow Q=4440\ W[/tex]
The rate of energy transfer is 4440 W