# Last one

1. May 2, 2005

### Jayhawk1

A copper rod and an aluminum rod of the same length and cross-sectional area are attached end to end. The copper end is placed in a furnace which is maintained at a constant temperature of 279oC. The aluminum end is placed in an ice bath held at constant temperature of 0.0oC. Calculate the temperature (in degrees Celsius) at the point where the two rods are joined. The thermal conductivity of copper is 380 J/(s m Co) and that of aluminum is 200 J/(s m Co).

2. May 2, 2005

### Andrew Mason

Since all points on the rod are at thermal equilibrium, the heat flowing in from the heat source is equal to the heat flowing out toward the ice.

For copper:

$$\frac{dQ_{in}}{dt} = \lambda_{cu}A\frac{dT}{dx}$$

For aluminum:

$$\frac{dQ_{out}}{dt} = \lambda_{al}A\frac{dT}{dx}}$$

Since heat in = heat out:

$$\lambda_{cu}\frac{dT}{dx} = - \lambda_{al}\frac{dT}{dx}$$

Now for copper:

$$\frac{dT}{dx} = (T_{hot} - T_{mid})/L$$

and for Aluminum:

$$\frac{dT}{dx} = (T_{mid} - T_{cold})/L$$

Solve for the middle temp.

AM

Last edited: May 2, 2005