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Question

0.1 kg of a substance absorbs 1200 J of heat, increasing its temperature from 20°C to 50°C. What is its specific heat capacity?

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Explanation

**First law applications** for isobaric W = P ΔV, Q = n C_p ΔT, ΔU = n C_v ΔT, for isothermal ideal gas ΔU=0 Q=W=n R T ln(V₂/V₁), for adiabatic Q=0 W= -ΔU = (P₁V₁ - P₂V₂)/(γ-1), for isochoric W=0 ΔU=Q=n C_v ΔT, enabling calculation of Q,W,ΔU for any process. Specific heat: s = (Δ Q)/(m Δ T) . Given Δ Q = 1200 J , m = 0.1 kg , Δ T = 50 - 20 = 30 K . s = (1200)/(0.1 × 30) = 400 J kg⁻¹ K⁻¹ . Using first law ΔU = Q - W, W = ∫ P dV,

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