Find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C. The specific heat capacity of water is 4200 J/kg-K and its densities at 0°C and 4°C are 999.9 kg/m3 and 1000 kg/m3 respectively. Atmospheric pressure = 105 Pa.

Sol. Given M = 2 kg 2t = 4°c Sw = 4200 J/Kg–k f0 = 999.9 kg/m^3 f4 = 1000 kg/m^3 P = 10^5 Pa. Net internal energy = dv dQ = DU + dw ⇒ ms∆Q∅ = dU + P(v^0 – v^4) ⇒ 2 × 4200 × 4 = dU + 10^5(m – m) ⇒ 33600 = dU + 10^5 (m/V base 0 – m/v base 4) = dU + 10^5(0.0020002 – 0.002) = dU + 10^5 0.0000002 ⇒ 33600 = du + 0.02 ⇒ du = (33600 – 0.02) J