A wall is composed of two materials. Material 1 has a thickness of 4.81 cm and a
ID: 2251768 • Letter: A
Question
A wall is composed of two materials. Material 1 has a thickness of 4.81 cm and a thermal conductivity of 0.1 while material 2 has a thickness of 5.13 cm and a conductivity of 1. If the temperature difference inside to outside is 25 Co and the wall has an area of 10 m2, what is the energy loss per second to the nearest watt?
If the energy loss is 553 watts, the material 2 has a thickness of 5.87 cm and the inside temperature is 25.8 oC, what is the temperature of the interface to the nearest tenth of a degree Celsius?
A wall is composed of two materials. Material 1 has a thickness of 4.81 cm and a thermal conductivity of 0.1 while material 2 has a thickness of 5.13 cm and a conductivity of 1. If the temperature difference inside to outside is 25 Co and the wall has an area of 10 m2, what is the energy loss per second to the nearest watt?Explanation / Answer
U = 1 / ( 0.0481/0.1 + 0.0587/1 )
U = 1.8528 W/m^2*C
Now use Newton's Law of Cooling in the form of:
qx = 1.8528*10*(T-25.8) = 553
T - 25.8 = 29.84
T = 55.64
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