Concepts Of Physics MCQ Edition [Volume 2]PhysicsHeat Transfer
A rod of length 20 cm has one end placed inside a furnace at 800 K . The sides of the rod are covered by an insulating material, whereas the other end behaves as a blackbody and emits radiation. During the steady state, the temperature of this exposed end is 750 K . The ambient air temperature is 300 K . Considering radiation as the sole significant method of energy transfer between the exposed end of the rod and the
Options
- A76 W m ⁻¹ K ⁻¹
- B10 W m ⁻¹ K ⁻¹
- C37 W m ⁻¹ K ⁻¹
- D74 W m ⁻¹ K ⁻¹
Correct answer
D. 74 W m ⁻¹ K ⁻¹
Step-by-step solution
In steady state, the rate of heat conduction through the rod is equal to the net rate of heat radiated by the exposed end. KA(T₁ - T₂) L = A (T₂^4 - T₀^4) K = L (T₂^4 - T₀^4) T₁ - T₂ Substituting the given values: = 6.0 10⁻⁸ W m ⁻² K ⁻⁴ L = 20 cm = 0.2 m T₁ = 800 K T₂ = 750 K T₀ = 300 K K = 6.0 10⁻⁸ 0.2 (750^4 - 300^4) 800 - 750 K = 1.2 10⁻⁸ (3164.0625 10^8 - 81 10^8) 50 K = 1.2 3083.0625 50 K = 3699.675 50 = 73.9935 W m ⁻¹ K ⁻¹ Rounding to the nearest integer, we get K 74 W m ⁻¹ K ⁻¹ .