Thermodynamics Question heat through a wall

[kdinser]

I'm stuck on this Thermo question dealing with heat transfer through a "plain wall"

They give us the area of the transfer A = 1 m^2
They give us the thickness of the wall L = .2 m
and they give us the rate of steady state energy transfer rate Qx = .15 kW

They then ask, if the temperature distribution is linear, what is the temperature difference across the wall in K.

I'm not sure how to get going on this one. Because they give actual numbers, I'm sure they want a real value for the number of Kalvins difference between the temperatures. I'm assuming they want me to use Fourier's Law here, but they don't give enough info to solve for (T2 - T1) directly. They don't give the value for the thermal conductivity of the wall, unless there is some standard value I'm supposed to use for a "plain wall". I seem to remember there was some kind of algebraic trick that I used to use to solve similar problems in physics, but the details escape my memory at the moment. If someone could get me started with this problem or give me hint to jog my memory, I'd appreciate it.

 

[mukundpa]

I too need the coefficient of thermal conductivity!

 

[quicknote]

Hello, it seems like you are on the right track with using Fourier's Law to solve this problem. You are correct in saying that you need the thermal conductivity of the wall in order to find the temperature difference across the wall. However, you can use a standard value for the thermal conductivity of a "plain wall" if it is not specified. In this case, you can use a value of 0.04 W/mK for the thermal conductivity of the wall.

To solve for the temperature difference, you can use the formula Q = kA(T2-T1)/L, where Q is the steady state energy transfer rate, k is the thermal conductivity, A is the area of transfer, and L is the thickness of the wall.

Plugging in the given values, you can rearrange the equation to solve for (T2-T1). This will give you the temperature difference in Kelvin. Remember to convert the given value of 0.15 kW to W before plugging it into the equation.

I hope this helps you solve the problem. If you need further assistance, please let me know. Good luck!