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calculate the increase in internal energy of 1 kg of water at `100^(o)C` when it is converted into steam at the same temperature and at 1atm (100 kPa). The density of water and steam are 1000 kg `m^(-3)` and 0.6 kg `m^(-3)` respectively. The latent heat of vaporization of water `=2.25xx10^(6) J kg^(1)`.

Text Solution

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The volume of 1 kg of water
`=(1)/(100)m^(3)` and of 1 kg of steam `=(1)/(0.6)m^(3)`. The increase in volume
`=(1)/(0.6)m^(3)-(1)/(1000)m^(3)`
`=(1.7-0.001)m^(3)~~1.7m^(3)`.
The work done by the system is `pDeltaV`
`=(100 kPa)(1.7 m^(3))`
`=1.7xx10^(5) J`.
the heat given to convert 1 kg of water into steam
`2.25xx10^(6) J`.
the change in internal energy is `DeltaU=DeltaQ-DeltaW`
`=2.25xx10^(6) J-1.7xx10^(5) J`
`=2.08xx10^(6) J`.
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