Temperature at the junction of 2 wires

[Deebu R]

Homework Statement

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?

Homework 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?

 

[BvU]

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

 

[Deebu R]

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

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

 

[BvU]

Something like Fourier's law, perhaps ?

 

[Deebu R]

Something like Fourier's law, perhaps ?

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?

 

[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.

 

[Deebu R]

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.

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?

 

[BvU]

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

 

[gneill]

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:

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