A heat flux of 4000/J is to be passed through a copper rod of length 10cm and area of cross - Physics - Electrostatic Potential And Capacitance

Question

A heat flux of 4000/J is to be passed through a copper rod of
length 10cm and area of cross section 100sq cm The thermal
conductivity of copper is 400W/mC The two ends of his rods must be
kept ta temperature difference of?

Sakshi Gupta · Accepted Answer

Dear student,

Using dQ/dt = KAâˆ†T/l

we get âˆ†T = l/(K x A) x dQ/dt = 0 1/(400 x 100 x
10-4) x 4000 = 100oC

Hope this helps you
Regards

A heat flux of 4000/J is to be passed through a copper rod of length 10cm and area of cross section 100sq. cm. The thermal conductivity of copper is 400W/mC . The two ends of his rods must be kept ta temperature difference of?