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1 "cm"^3 of water at 1 atm pressure is c...

`1 "cm"^3` of water at 1 atm pressure is converted into steam, volume of steam becomes `1671 "cm"^3` hence increase in internal energy will be ____

A

2087 J

B

167 J

C

373 cal

D

373 J

Text Solution

Verified by Experts

The correct Answer is:
C
`DeltaQ=mL_V`
=1 x 540
=540 cal
Now from the first law of thermodynamics,
`DeltaQ=DeltaU+DeltaW`
`DeltaQ=DeltaU+PDeltaV`
`therefore DeltaU=DeltaQ-PDeltaV`
`=540-10^5(1671-1)xx10^(-6)`
`=540-1670xx10^(-1)`
=540-167
=373 cal
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