1 kg of water boils at 373 K at a constant pressure of 10^5 Pa . The specific volume of steam is 1.6 m ^3/ kg and the specific volume of liquid water is negligible. If the latent heat of vaporization of water is 2260 kJ/kg , the change in internal energy of the system during this process is :

Step-by-step solution

Heat absorbed by the water during boiling is Q = mL_ v = 1 2260 = 2260 kJ . The change in volume is V = m(v_ s - v_ w ) = 1 (1.6 - 0) = 1.6 m ³ . Work done by the system against the constant atmospheric pressure is W = P V = 10⁵ 1.6 = 1.6 10⁵ J = 160 kJ . According to the first law of thermodynamics, Q = U + W . Therefore, the change in internal energy is U = Q - W = 2260 - 160 = 2100 kJ . Answer: 2100 kJ