# Homework Help: Thermodynamics - conduction

1. Mar 5, 2014

### Feodalherren

1. The problem statement, all variables and given/known data
A bar of gold (Au) is in thermal contact with a bar of
silver (Ag) of the same length and area (Fig. P20.39).
One end of the compound bar is maintained at 80.0°C,
and the opposite end is at 30.0°C. When the energy
transfer reaches steady state, what is the temperature at
the junction?

The figure just shows that the two blocks are insulated from all other directions - so ignore conduction to anything else.
2. Relevant equations

P=kA(ΔT)/L

3. The attempt at a solution
If one side is maintained at 80C wouldn't the whole system just keep warming up until it reaches 80C (steady state) then just conduct the same amount of heat that comes in from one side to the other side?

2. Mar 5, 2014

### Staff: Mentor

How does the system know that it's supposed to be 80 C and not 30C? If the would system reaches 80 C, how is the heat supposed to be conducted from one side to the other side with no temperature gradient present?

Let T be the temperature at the interface. What is the rate of heat conduction from the 80 C boundary to the interface? What is the rate of heat conduction from the interface at temperature T to the boundary at 30 C? What is the relationship between these two rates of heat conduction at steady state?

Chet

3. Mar 5, 2014

### Feodalherren

See that's what throws me off. The problem says that one side is maintained at 80C, it says nothing about the other side being maintained. If one side is maintained and the other is not, doesn't the whole system go to equilibrium at whatever temperature is maintained at one side? So once both the Ag and Au are at 80C it conducts heat to the 30C side and keeps doing that until it also goes to 80C.

If I have a metal rod and I keep it on a flame the metal rod will eventually become as hot as the fire (ignoring any other sources of conduction, radiation or convection).

But going with what you said I'd do something like this

$K_{Au}A\frac{T-80}{L} = K_{Ag}A\frac{T-30}{L}$

4. Mar 6, 2014

### Staff: Mentor

The problem statement meant to imply that the other end was maintained at 30 C. They didn't intend for you to be solving a transient heat transfer problem.
This result is correct (except for a sign error), and is all that they were looking for.

Chet

5. Mar 6, 2014

### Ronaldo95163

It's 80-T on the left and T-30 on the right but otherwise that equation is correct :D