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Question

0.25 kg of a substance at 30°C absorbs 1500 J of heat, reaching 60°C. What is its specific heat capacity?

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Explanation

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. Specific heat: s = (Δ Q)/(m Δ T) . Δ Q = 1500 J , m = 0.25 kg , Δ T = 60 - 30 = 30 K . s = (1500)/(0.25 × 30) = 200 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

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