Question

What is the result of mixing 1kg ice at 0°c with 100kg steam at 100°c? Solve

Answer 1

Compare Latent heat of fusion of ice`=3.36 × 10^5\ Jcdotkg^-1`
and
Latent heat of vaporization of water `= 2.26 × 10^6\ Jcdotkg^-1`

We see that steam will condense to become water at `100^@C` and Ice will melt to become water at `0^@C`

Heat required to melt `1\ kg` of ice to become water at `0^@C`
`=3.36 × 10^5\ J`
This much heat can be given by amount of steam`=(3.36xx10^5)/(2.26xx10^6)`
`=0.151\ kg`

Heat required to raise temperature of `1\ kg` of water at `0^@C` to `100^@C` `=msDeltaT=1xx4186xx100=4.186xx10^5\ J`
(Specific heat of water `s=4186\ J cdot kg^-1 C^-1`)

This amount of heat can be provided by condensing of steam`=(4.186xx10^5)/(2.26 × 10^6)=0.185\ kg`

Total steam condensed `=0.151+0.185=0.336\ kg`

Hence in the final mixture we will have
water at `100^@C=1.336\ kg` and steam `99.664\ kg` of steam