Answer to Question #325591 in Physics for Naruto

Question #325591

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
2022-04-11T12:05:35-0400

The Fourier's law says

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

Hence, the temperature difference

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

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

The temperature at the other surface

"T_2=T_1\\pm\\Delta T=40.0\\pm3.33\\\\\n=43.3^\\circ\\rm C \\: \\rm or\\; 36.7^\\circ\\rm C"


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