How much heat is required to raise the temperature of 0.6 kg of silver from 20^circ C to 50^circ C ? (Specific heat of s
**Cyclic work** equals area inside loop, for rectangular cycle P₁V₁→P₂V₁→P₂V₂→P₁V₂→P₁V₁, W = (P₂-P₁)(V₂-V₁), heat absorbed and rejected during different legs, net work output for heat engine, input for refrigerator. Δ Q = m s Δ T . m = 0.6 , s = 236.1 , Δ T = 50 - 20 = 30 . Δ Q = 0.6 × 236.1 × 30 = 4249.8 J ≈ 4250 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h,
Ref: NCERT > Physics Book > Thermodynamics > Cyclic Processes and Reversibility Concepts