(adsbygoogle = window.adsbygoogle || []).push({}); [SOLVED] Heat equilibrium between two different rods

1. The problem statement, all variables and given/known data

A rod 1.3 m long consists of a 0.8 m length of Aluminium joined end-to-end to a 0.5 m length of brass. The free end of the aluminium section is maintained at 150 oC and the free end of the brass piece is maintained at 20 oC. No heat is lost through the sides of the rod. In the steady state, what is the temperature of the point where the two metals are joined ? (Thermal conductivity of brass k_b = 109 W/mK and Thermal conductivity of Aluminum k_a = 237 W/mK).

2. Relevant equations

[tex]H=-kA[/tex] [tex]\frac{\Delta T}{\Delta x}[/tex]

3. The attempt at a solution

Well, you know in steady state that there is no net heat flow so H_aluminum = H_brass. So you have...(I give up for tying to figure this formatting out for now so please forgive me)

[tex](-k_{a} A (T_{h} - T_{e})) / l_{a} = (-k_{b} A (T_{e} - T_{c})) / l_{b}[/tex] , with [tex]T_{e}[/tex] being the point in the middle and [tex]T_{h}[/tex] and [tex]T_{c}[/tex] being the hold and cold ends respectively.

Since the cross sectional area of the two are the same, those will cancel and youre left with...

[tex](-k_{a} T_{h} + k_{a} T_{e}) / l_{a} = (-k_{b} T_{e} + k_{b} T_{c}) / l_{b}[/tex]

solving for [tex]T_{e}[/tex] you get...

[tex]k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a} = k_{b} T_{e} l_{a} + k_{a} T_{e} l_{b}[/tex]

Which then goes to...

[tex]T_{e} = (k_{a} T_{h} l_{b} + k_{b} T_{c} l_{a}) / (k_{b} l_{a} + k_{a} l_{b})[/tex]

However, when I put in the values I have, I get an answer HIGHER than the original [tex]T_{h}[/tex]!

I dont get what Im doing wrong here.

I have a problem after this where I have to solve for [tex]T_{e}[/tex] when 3 rods come together but I clearly cant do that one properly if I cant solve this one first!

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# Homework Help: Heat equilibrium between two different rods

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