1 kg of water at 100^ C is converted to steam at the same temperature. Volume of 1 cc of water changes to 1671 10^3 cc on boiling. The change in internal energy of the system is (Latent Heat of vaporisation of water is 22.68 10^5 J kg⁻¹ ; 1 atm = 1.0 10^5 Pa)

Step-by-step solution

Given the final volume of 1 kg of steam is 1671 10^3 cc = 1.671 m ^3 . Initial volume of 1 kg of water is 1000 cc = 10⁻³ m ^3 . Change in volume, V = 1.671 - 0.001 = 1.670 m ^3 . Work done, W = P V = 1.0 10^5 1.670 = 1.67 10^5 J. Heat supplied, Q = mL = 1 22.68 10^5 = 22.68 10^5 J. Using the first law of thermodynamics, Q = U + W . U = Q - W = 22.68 10^5 - 1.67 10^5 = 21.01 10^5 J. Answer: 21.01 10^5 J