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I'm having trouble solving this question where two metal plates are in thermal contact with each-other. I'm given the type of metals in each plate, the thickness and the overall temperature difference for both plate however I don't understand how to find the temperature difference over a specific metal plate.

so far I've tried applying the heat conduction formula:

$ P_{cond}= \frac{kA\Delta T}{L} $

and equating it for the other material

$ \frac{k_{Al}\Delta T}{L_{Al}} = \frac{k_{Pb}\Delta T}{L_{Pb}} $

I know that $\Delta T$ is the temperature difference of that specific plate, however I'm finding it hard to understand how to actually find this given the overall difference.

Any help would be greatly appreciated.

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You're nearly there! You've worked out that:

$$ \frac{k_{Al}\Delta T_{Al}}{L_{Al}} = \frac{k_{Pb}\Delta T_{Pb}}{L_{Pb}} $$

where $\Delta T_{Al}$ is the temperature drop in the aluminium plate and $\Delta T_{Pb}$ is the temperature drop in the lead plate. Now just use the fact that the sum of the two temperature drops is the total temperature drop:

$$ \Delta T_{Al} + \Delta T_{Pb} = \Delta T $$

Now you have two simultaneous equations for the two variables $\Delta T_{Al}$ and $\Delta T_{Pb}$. Just solve the equations.

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  • $\begingroup$ So if I were to separate the temps as $T_1, T_2 and T_3$, $\Delta T_{overall}$ would equal $T_1-T_3$ right? $\endgroup$
    – Daniel Munoz
    Commented Apr 7, 2016 at 6:18
  • $\begingroup$ Wait! like this? $\frac{k_{Al}L_{Pb}}{L_{Al}k_{Pb}} \Delta T_{Al} =\Delta T_{Pb} $ Then sub into the sum of $\Delta T_{overall}$ $\endgroup$
    – Daniel Munoz
    Commented Apr 7, 2016 at 6:27

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