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1cm³of water at its boiling point absorbs 540 calories of heat to become steam with a volume of 1671 cm³ .If the atmospheric pressure = 1.013x10⁵ N / m² and the mechanical equivalent of heat = 4.19 J / calorie , the energy spent in this process in overcoming intermolecular forces is

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540 cal

40 cal

500 cal

Zero

answer is C.

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Detailed Solution

$\Delta Q = \Delta U + \Delta W$
$\therefore \Delta U = \Delta Q - \Delta W = 540 - \frac{{P({V_2}-{V_1})}}{J}$
$= 540 - \frac{{1.013 \times {{10}^5} \times [(1671 - 1) \times {{10}^{ - 6}}]}}{{4.2}}$
$= 540 - 39.7 = 500\;calories$

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