# Temperature at the junction of 2 wires

1. Sep 14, 2016

### Deebu R

1. The problem statement, all variables and given/known data
The thermal conductivity of copper is 4 times that of brass.Two rods of copper and brass having same length and cross section are joined end to end. The free end of copper is at 0 degree C and the free end of brass is at 100 degree C. The temperature at the junction is?

2. Relevant equations
I don't know what equation to use.
R= r l/A?
Or thermal conductance = kl/A where k is the thermal conductivity?
Or is it something else?

2. Sep 14, 2016

### BvU

Hi,
do you have a relevant equation that links temperature (or temperature difference) and thermal conductivity ?

3. Sep 14, 2016

### Deebu R

Sorry but I got nothing. Any hint or clue to help?

4. Sep 14, 2016

### BvU

5. Sep 14, 2016

### Deebu R

To be honest this is the first time I have heard about that law.
So I got like,

Q=-k (0-100)= 100k
Q=-4k(100-0)= -400k?

6. Sep 14, 2016

### BvU

That wouldn't work: the junction temperature doesn't appear. But you are on the right track. Length and area should indeed divide out. With heat and temperature drop it's just like with current and voltage drop. Check the dimensions.

7. Sep 15, 2016

### Deebu R

I think I got it ,
Thermal conductivity of brass=k
Thermal conductivityof copper=4k
Area of rods= A and length=l
Temperature of the junction = θ
Rate of flow of heat= KA(100-θ)/l = 4kA(0-θ)/l

(100-θ)=4(0-θ)
5θ=100
θ= 20° C.
Correct?

Last edited: Sep 15, 2016
8. Sep 15, 2016

### BvU

Yes. Make it a habit to add (and check) dimensions.

9. Sep 15, 2016

### Staff: Mentor

That works. You've reach the correct result.

Note that heat flow and the flow of electricity are analogous. In effect we can analyze a thermal "circuit" as we would an electric circuit: Temperature difference is analogous to potential difference, heat flow to current flow, thermal conductivity to electrical conductivity. In either domain conductance is the reciprocal of resistance.

In this problem the two rods are identical except for material so we can immediately conclude that the brass rod will have 4 times the thermal resistance as the copper rod. The thermal circuit is analogous to a potential divider:

And since in a potential divider it's the resistance ratio that's important:

and you can write:

$T_j = 100~C \cdot \frac{1}{1 + 4} = 20~C$