STANDARDPhysics-Thermodynamics

STANDARD Physics Heat Transfer MCQ Question

Type: MCQ-numerical-Medium-Class 11

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

A

120 W m⁻²

B

90 W m⁻²

C

200 W m⁻²

D

65 W m⁻²

Correct Answer

Option B

Detailed Explanation

The energy flux is given by k(T1T2)d\frac{k(T_1 - T_2)}{d}, where k is the thermal conductivity, T₁ and T₂ are the temperatures, and d is the thickness. Substituting the given values, 0.1×(1000100)1=90\frac{0.1 \times (1000 - 100)}{1} = 90 W m⁻².

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