The temperature θ at the junction of two insulating sheets, having thermal resistances R₁ and R₂ as well as top and bottom temperatures θ₁ and θ₂ (as shown in figure) is given by :

Solution

Physics sir · 4 steps

Two sheets in series carry the same heat current, so equate the flow through R₂ and through R₁.

1. Same heat current. In steady state no heat piles up at the junction. The heat per second entering it through one sheet equals the heat per second leaving through the other. A thermal resistance works like an electrical one: heat current = Δθ / R.

   (θ₂ - θ)/R₂ = (θ - θ₁)/R₁

2. Solve for the junction. Cross-multiply to get R₁(θ₂ - θ) = R₂(θ - θ₁). Collect the θ terms: θ(R₁ + R₂) = θ₁ R₂ + θ₂ R₁.
3. Quick check. If R₂ is very large, almost all the temperature drop happens across R₂, so θ should be close to θ₁. The result gives θ → θ₁ as R₂ → ∞, as expected.
4. So. θ = (θ₁ R₂ + θ₂ R₁)/(R₁ + R₂), which is option (a).

The idea to keep. For slabs in series, the heat current is the same through each. Write Δθ/R for each slab and equate them. Each end temperature gets weighted by the resistance of the far sheet, not its own.

Answer from the NTA final answer key. Solution by PrepWiser.