Q: 1 cm3 of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm3. If the atmospheric pressure = 1.013 ×105 N/m2 and the mechanical equivalent of heat = 4.19 J/cal., the energy spent in this process in overcoming intermolecular forces is
(a) 540 cals.
(b) 40 cals.
(c) 500 cals.
(d) zero
Ans: (c)
Sol: According to First Law of Thermodynamics
dQ = dU + dW
dU = dQ-dW
= mL – P(dV)
$ \displaystyle 1\times 540 – \frac{1.013 \times 10^5 (1671 – 1)\times 10^{-6}}{4.2} $
= 540-40
= 500 cals