1. Tardigrade
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  3. Physics
  4. 1 cm 3 of water at its boiling point absorbs 540 cal of heat to become steam with a volume of 1671 cm 3 the atmospheric pressure =1.013 × 105 Nm -2 and the mechanical equivalent of heat =4.19 cal, the energy spent in this process in overcoming intermolecular forces is:

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Q. $1 cm ^{3}$ of water at its boiling point absorbs 540 cal of heat to become steam with a volume of $1671 cm ^{3}$ the atmospheric pressure $=1.013 \times 10^{5} Nm ^{-2}$ and the mechanical equivalent of heat $=4.19$ cal, the energy spent in this process in overcoming intermolecular forces is:

Thermodynamics

Solution:

Work done by steam $= P \Delta V$
$=1.013 \times 10^{5}[1671-1] \times 10^{-6}$
$=169.17$ Joule
$=41.06$ cal $\therefore [1$ cal $=4.2$ Joule $]$
energy spent to overcoming intermolecular forces is
$=540-41.06=498.94 cal \simeq 499 cal$