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1 cm^(3) of water at its boiling point a...

`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

A

540 cal

B

40 cal

C

500 cal

D

Zero

Text Solution

Verified by Experts

The correct Answer is:
C
Since, `DeltaQ=DeltaU + DeltaW`
`implies DeltaU=540-(P(V_2-V_1))/(4.2 J)`
` implies DeltaU=540-(1.013xx10^5 xx(1671-1)xx10^(-6))/(4.2)`
`implies DeltaU` = 540-40 = 500 cal.
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