# What is the temperature at the joint between the two rods?

1. Jun 14, 2010

### mlostrac

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

The thermal conductivity of aluminum is twice that of brass. Two rods, one of
aluminum and the other of brass, are joined together end to end in excellent
thermal contact. They are of equal lengths and radii. The free end of the brass
rod is maintained at 0°C while the free end of the aluminum rod is kept at 200°C.
Under these conditions of steady heat flow, what is the temperature at the joint
between the two rods?

2. Relevant equations

Q/t = KA (T1-T2)/L

3. The attempt at a solution

I was thinking I had to set the rate of heat flow in the brass and aluminum to be equal and then isolate to find T1, but that turned out to be way wrong.

How should I start? I read a similar problem that said we should add the conductivities of the two metals. For example, conductivity of brass = x, conductivity of aluminum = 2x, therefore overall conductivity is 3x.

2. Jun 15, 2010

### hikaru1221

I think your idea to set the heat flow rates to be equal is correct. I'm not sure why you got wrong answer. Did you notice that for the brass rod, T is the temperature of the hotter end and for the aluminum rod, it's that of the cooler end, where T is the temperature at the joint (and so, T is at the place of T1 for one rod but at the place of T2 for the other rod)?

3. Jun 15, 2010

### Mapes

This sounds like solid reasoning. Where did it go wrong? Did you write a conduction equation for each metal and set them equal?

Not for two materials connected end to end.

4. Jun 16, 2010

### mlostrac

K heres what I did:

[kA (T1-T2)]/L = [2kA (T1-T2)]/L

Since Areas and lengths are the same, the A's and L's cancel:

k(T1-T2) = 2k(T1-T2)
k(273 - T2) = 2k(473 - T2)

273k - KT2 = 946k - 2kT2
kT2 = 673k

T2 = 673 Kelvin or 400 degrees celsius

Which has got to be wrong because the original temp of the aluminum (the hotter material) is 200 degrees celsius

What do you think?

5. Jun 16, 2010

### Mapes

I don't know where the 2k came from; if aluminum has twice the thermal conductivity, why multiply it by two again? Try setting an energy balance: the heat flow coming in equals the heat flow going out.

Assume the aluminum rod is on the left, and the temperatures at measured at the left end, the midpoint, and right end ($T_\mathrm{l}$, $T_\mathrm{m}$, and $T_\mathrm{r}$). The heat flow is the positive x direction is always

$$-kA\frac{\Delta T}{\Delta x}=-k_\mathrm{Al}A\frac{T_\mathrm{m}-T_\mathrm{l}}{L}=-k_\mathrm{Br}A\frac{T_\mathrm{r}-T_\mathrm{m}}{L}$$

Try solving this.

6. Jun 16, 2010

### mlostrac

Thermal conductivity of aluminum (2k) is 2 times that of brass (k). Thank you, I'll try now...

7. Jun 16, 2010

### Mapes

If a person with height X is twice as tall as a person with height Y, does Y = 2X?

8. Jun 16, 2010

### mlostrac

OK, so I ended up getting 133 degrees celsius after using your equation which sounds like it could be right. I used the thermal conductivity values found in my text to plug into the following equation:

Tm = (Kal473 + Kbr273) / (Kal + Kbr)

* the numbers for Kal and Kbr were 200 and 100 respectively.

Look good?

9. Jun 16, 2010

### Mapes

That sounds better. It should make sense that if aluminum's thermal conductivity is higher, then the midpoint should be closer to the temperature at the end of the aluminum rod.

10. Jun 16, 2010

### mlostrac

Ok thank you for your help! Much appreciated