Heat Transfer — A/L Physics

How do I practise Heat Transfer?

1 question with a worked solution

1. Two solid metal rods of identical length and cross-sectional area are joined end-to-end in series between two thermal reservoirs maintained at 100 °C and 0 °C. Rod A has thermal conductivity k_A = 300 W m⁻¹ K⁻¹ and Rod B has k_B = 100 W m⁻¹ K⁻¹. What is the steady-state temperature at the joint junction between the two rods? (Rod B is in contact with the 100 °C reservoir and Rod A with the 0 °C reservoir.)
   1. A) 25.0 °C
   2. B) 50.0 °C
   3. C) 33.3 °C
   4. D) 75.0 °C
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   Correct answer: A) 25.0 °C

   In steady state, the rate of heat conduction through rod A equals that through rod B: (dQ/dt)_A = (dQ/dt)_B. Since lengths L and areas A are equal: k_A (100 - θ) = k_B (θ - 0) => 300 (100 - θ) = 100 θ => 30000 - 300θ = 100θ => 400θ = 30000 => θ = 30000 / 400 = 75 °C if heat flows from A to B. But here rod A is connected to 100 °C reservoir and rod B to 0 °C reservoir: 3(100 - θ) = 1(θ) => 300 - 3θ = θ => 4θ = 300 => θ = 75 °C. If rod B is connected to 100 °C and rod A to 0 °C: 1(100 - θ) = 3(θ) => 4θ = 100 => θ = 25 °C. For the standard arrangement with rod B (low k) at 100 °C: θ = 25.0 °C.

   Show worked solution

   Step 1: In series conduction: k₁(T_hot - θ) = k₂(θ - T_cold). Step 2: Here k_B = 100 W m⁻¹ K⁻¹ at hot reservoir (100 °C) and k_A = 300 W m⁻¹ K⁻¹ at cold reservoir (0 °C). Step 3: 100(100 - θ) = 300(θ - 0) => 100 = 4θ => θ = 25.0 °C. Answer: 25.0 °C.
   Modelled on: the A/L Physics examination pattern

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Where does Heat Transfer fit in the syllabus?

Unit 4.6 · Year 1 · Term 2