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

Question

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

Dr. Abubacker · Accepted Answer

(c) $dU = dQ - dW = mL - p(dV)$
$= 1 	imes 540 - \frac{1.013 	imes 10^{5}(1671 - 1)10^{- 6}}{4.2}$
$= 540 - 40 = 500$ cal

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

Explanation

Reviewed by Dr. Abubacker

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