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1.0 m^(3) of water is converted into 167...

`1.0 m^(3)` of water is converted into `1671 m^(3)` of steam at atmospheric pressure and `100^(@)C` temperature. The latent heat of vaporisation of water is `2.3xx10^(6) J kg^(-1)`. If 2.0 kg of water be converted into steam at atmospheric pressure and `100^(@)C` temperature, then how much will be the increases in its internal energy? Density of water `1.0xx10^(3) kg m^(-3)`, atmospheric pressure =`1.01xx10^(5) Nm^(-2)`.

Text Solution

Verified by Experts

Heat given to water to change into steam
`Q=ML=2.0xx2.3xx10^(6)`
`=4.6xx10^(6)J`
volume of water `V=("Mass")/("Density")=2.0/(10^(3))`
`=2.0xx10^(-3)m^(3)`
Volume of steam formed will be
`=2.0xx10^(-3)xx1671`
`=3342xx10^(-3)m^(3)`
The change in volume in the process
`DeltaV=V=3342xx10^(-3)-2.0xx10^(-3)`
`=3340xx10^(-3)m^(3)`
The work done against the atmospheric pressure
`W=P DeltaV`
`=(1.01xx10^(5))xx(3340xx10^(-3))`
`=0.337xx10^(6)J`
By first law the thrmodynamics
`Q=DeltaU+W`
`:.DeltaU=Q-W`
`=4.6xx10^(6)-0.337xx10^(6)`
`=4.263xx10^(6)J`
Here positive value of `DeltaU` indicates that internal energy in the process increases.
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