A heat source at T = 10 3   K is connected to another heat reservoir at T = 10 2   K by a copper…

A heat source at T=103 K is connected to another heat reservoir at T=102 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:
  1. 65 W m-2
  2. 120 W m-2
  3. 90 W m-2
  4. 200 W m-2

Solution

Given,

Temperature of heat source, TH=103 K,

Temperature of heat reservoir, TL=102 K,

Conductivity of copper K=0.1 W K-1m-1

From the equation of steady-state heat flow,

dQdt=KATL

Heat energy flux can be defined as the rate of heat energy transfer through a given surface.

1AdQdt=K×1000-1001=90 W m-2

Asked in: JEE Main 2019 (10 Jan Shift 1)

Practice more Thermal Properties of Matter questions on Aicharya