# Temperature profile at steady state [closed]

Consider a solid wall consisting of three layers where the layers are of equal thickness but made out of different materials: $$\ce{Layer 1}$$ is made out of copper (high thermal conductivity), $$\ce{Layer 2}$$ is made out of a highly porous ceramic (low thermal conductivity), and $$\ce{Layer 3}$$ is made out of concrete (medium thermal conductivity).

How does the steady state temperature profile inside the wall look like?

I'm not sure if this question is asked in a way to mislead me into drawing the T-profile as (2) when at steady-state it actually should look like (1).

I apologize if this question is seen as off-topic or simple, but I would appreciate it if someone would explain if I am correct in saying that the T-profile at steady state should be linear.

Thank you!

(1) (2)

## closed as off-topic by Mithoron, Soumik Das, Tyberius, Todd Minehardt, Jon CusterMar 13 at 22:26

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• No, the stationary profile won't be linear. Why would it? – Ivan Neretin Mar 12 at 12:28
• If you would provide your thoughts on why you think it might be one or the other it would be on topic. – A.K. Mar 13 at 17:43

No, the temperature gradient is not uniform across all three layers.

To see why, start from the Fourier law for heat flow, shown in my solution to this problem.

Since at steady state the heat flux $$q$$ must be equal for all layers, we have that

$$q_{1}=q_{2}=q_{3}$$

And since the layers are equal in thickness,

$$k_1\Delta{T_1}=k_2\Delta{T_2}=k_3\Delta{T_3}$$

Clearly, if the $$k$$ are not equal, the $$\Delta{T}$$ cannot be equal either, so the profile will not be that in (1).