Determine the increase in the internal energy of 10 g of water as it is heated from 0^ C to 100^ C and subsequently transformed into steam at 100 kPa . The density of the steam is 0.6 kg m ⁻³ . Assume the specific heat capacity of water is 4200 J kg ⁻¹ ^ C ⁻¹ and the latent heat of vaporization of water is 2.25 10^6 J kg ⁻¹ .

Step-by-step solution

Mass of water, m = 10 g = 0.01 kg Heat required to raise the temperature of water from 0^ C to 100^ C is given by: Q₁ = m c T = 0.01 4200 100 = 4200 J Heat required to convert water at 100^ C into steam is given by: Q₂ = mL = 0.01 2.25 10^6 = 22500 J Total heat supplied to the system is: Q = Q₁ + Q₂ = 4200 + 22500 = 26700 J Work done by the system against the constant external pressure during vaporization is: W = P(V_ steam - V_ water ) Neglecting the initial volume of water, the volume of steam is: V_ steam = m _