%3Cp%3E(1)%20%3Cspan%26nbsp%3Bstyle%3D%22color%3A%26nbsp%3Brgb(17%2C%26nbsp%3B17
ID: 2256916 • Letter: #
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steam quality = mass of steam / total mass
Initial mass of steam = 0.2*1.7 = 0.34 kg
Final mass = 1.7 kg
So, vapour formed = 1.7-0.34 = 1.36 kg = 1360 g = 1360/18 mole = 75.556 mole
So, At constant pressure, work done by gas = n*R*T = 75.556*8.314*373 = 234.306 kJ
{373 is boiling point, since, work is done by gas, it is negative according to sign convention}
Heat given to vapourize the 1.36 kg gas = m*Sv = 1.36*2260 kJ = 3073.6 kJ
U = Q + W = 3073.6 - 234.306 = 2839 kJ
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