Question

At 1 atmospheric pressure, 1.000 g of water having a volume of `1.000 cm^3` becomes 1671 `cm^3` of steam when boiled. The heat of vaporization of water at 1 atmosphere is `539 cal//g` . What is the change in internal energy during the process ?

Answer 1

Heat spent during vaporisation
Q = mL = `1.000 xx 539` = 539 cal
Work done `W = p(V_(v) - V_(l))`
`=1.013 xx 10^(5) xx (1671 - 1.000( xx 10^(-6)`
`169.2J = (169.2)/(4.18) cal = 40.5 cal`
`therefore` Change in internal energy,
U = 539 cal `- 40.5` cal
= 498.5 cal