Question

A uniform copper bar 100 cm long is insulated on side, and has its ends exposed to ice and steam respectively. If there is a layer of water 0.1 mm thick at each end, calculate the temperature gradient in the bar. `K_(Cu)=1.04` and `K_(water)=0.0014` in CGS units.

Answer 1

Let `theta_1` and `theta_2` be the temperatures at the ends of the copper bar.
Heat transfer per second through the system is
`(dH)/(dt)=(A(100-0))/(((0.001)/(K_W)+(100)/(K_(cu))+(0.01)/(K_W)))`
Heat transfer per second through copper bar
`=(K_(Cu)A(theta_1-theta_2))/(100)`
As the rods are in series, heat transfer per second must be same through each part.
`(A(100-0))/(((0.01)/(K_(W))+(100)/(K_(Cu))+(0.01)/(K_(W))))=(K_(Cu)A(theta_(1)-theta_(2)))/(100)`
putting `K_(Cu)=1.04` and `K_W=0.0014` we get
Temperature gradient`=((theta_1-theta_2))/(100)`
`=0.87^@Ccm^-1`