`1 cm^(3)` of water at its boiling point absorbs `540 cal` of heat to become steam with a volume of `1671 cm^(3)`. If the atmospheic pressure is `1.013 xx 10^(5) N//m^(2)` and the mechanical equivalent of heat `= 4.19 J//cal`, the energy spent in this process in overcoming intermolecular forces is

Options

A. 540 cal
B. 40 cal
C. 500 cal
D. zero

Answer

C

Explanation

As, from the first Thermodynamic Law
`dU=dQ-dW=mL-p(dV)`
= `1xx540-(1.013xx10^(5)(1671-1)10^(-6))/(4.2)`
= `540-40=500"cal"`