Question

A bar of gold (Au) is in thermal contact with a bar of silver (Ag) of the same length and area. One end of the compound bar is maintained at 80.0°C, Insulation and the opposite end is at 30.0°C.

Find the temperature at the junction when the energy flow reaches a steady state.

kgold = 314 J/(m*s*K)

ksilver = 427 J/(m*s*K)

Answer 1

Answer:- Temperature at the junction:

Explanation:-

Given:

• Temperature at hot junction of the compound bar :
• Temperature at cold junction of the compound bar : ​​​​​​​
• Thermal conductivity of gold :
• Thermal conductivity of silver :

The expression for the rate of energy transfer is given by,

At steady state, the rate of energy transfer from hot end is equal to the rate of energy transfer to the cold end, therefore,

Where,

• : is the temperature at the junction
• and : are the area of the gold and silver bar respectively
• and : are the lengths of the gold and silver bar respectively

Since the area and lengths of both the rods are same we have,

Therefore,

This is the temperature at the junction when the energy flow is at steady state.