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chapter :- Thermodynamics .
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Answer:
θ = 9θ ₂ /10 + θ₁ /10
Explanation:
Let interface temprature in steady state conduction is θ , then assuming no heat loss through sides ;
Rate of best flow through first slab = rate of best flow through second slab
→ 3K A (θ₂ - θ ) /d = KA (θ - θ₁)/3d
→ 9(θ₂ - θ ) = θ - θ₁
→ 9θ₂ + θ₁ = 10θ
→ θ = 9θ₂ / 10 + θ₁/10 Answer
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