calculate the increase in internal energy of 1 kg of water at `100(0)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)`.

calculate the increase in internal energy of 1 kg of water at `100(0)C

1.	calculate the increase in internal energy of 1 kg of water at `100(0)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)`.
Answer» 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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