Answer to Question #325443 in Physics for Chickennuggets

Question #325443

A wall with a thermal conductivity of 0.50W/m·K is maintained at 40.0°C. The heat


transfer through the wall is 250.0W. The wall surface area is 1.50m2 and its thickness is 1.00cm.


Determine the temperature at the other surface.

1
Expert's answer
2022-04-11T12:05:26-0400

The Fourier's law says

P=κΔTΔxAP=\kappa\frac{\Delta T}{\Delta x}A

Hence, the temperature difference

ΔT=PΔxκA\Delta T=\frac{P\Delta x}{\kappa A}

ΔT=250.00.010.501.50=3.33C\Delta T=\frac{250.0*0.01}{0.50*1.50}=3.33^\circ\rm C

The temperature at the other surface

T2=T1±ΔT=40.0±3.33=43.3Cor  36.7CT_2=T_1\pm\Delta T=40.0\pm3.33\\ =43.3^\circ\rm C \: \rm or\; 36.7^\circ\rm C


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