A heat source at T = 10 3 K is connected to another heat reservoir at T = 10 2 K by a copper slab which is 1 m thick. Given that the thermal conductivity of copper is 0 .1 W K - 1 m - 1 , the energy flux through it in the steady-state is:

Step-by-step solution

Given, Temperature of heat source, T H = 10 3   K , Temperature of heat reservoir, T L = 10 2   K , Conductivity of copper K = 0 . 1   W   K - 1 m - 1 From the equation of steady-state heat flow, d Q d t = K A ∆ T L Heat energy flux can be defined as the rate of heat energy transfer through a given surface. 1 A d Q d t = K × 1000 - 100 1 = 90   W   m - 2