Calculate the increase in the internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa. The density of steam = 0.6 kg/m3. Specific heat capacity of water = 4200 J/kg-C and the latent heat of vaporization of water = 2.25 × 106 J/kg.

Sol. Mass = 10g = 0.01kg. P = 10^5Pa dQ = Q base h2o 0° + Q base H2o – steam = 0.01 × 4200 × 100 + 0.01 × 2.5 × 10^6 = 4200 + 25000 = 29200 dW = P × ∆V ∆ = 0.01/0.6 – 0.01/1000 = 0.01699 dW = P∆V = 0.01699 × 10^5 1699J dQ = dW + dU or dU = dQ – dW = 29200 – 1699 = 27501 = 2.75 × 10^4 J