Question

A bar of lead (bar A) is in thermal contact with a bar of copper (bar B) of the same length and area. One end of the compound bar is maintained at Th = 82.5°C while the opposite end is at 30.0°C. Find the temperature at the junction when the energy flow reaches a steady state. _____ °C

Answer 1

Let us consider that the junction temperature is T.
Since the bar are in steady state therfore the heat transfer will be same from both bar.
heat transfer from bar A
Q₁ = k_AA(82.5 -T)/L
where k_A is thermal conductivity of bar A , A is area , L is length of the bar
Q₂ = k_bA(T-30) /L
where k_b is thermal conductivity of bar B
Now equating the heat transfer
Q₁ = Q₂
k_AA(82.5 -T)/L = k_bA(T-30) /L
k_A(82.5 - T) = k_b(T-30)
34.7*(82.5 -T) = 385*(T-30)
82.5 - T = 11.095*(T-30)
82.5 -T = 11.095T - 332.85
12.095T = 415.35
T = 34.34 ⁰C