Find the heat energy required to convert 10 g ice at `0^(@)C` to steam at `100^(@)C`. Specific latent heat of melting and vapourization are 336 KJ `kg^(-1)` and 2260 KJ `kg^(-1)`, respectively, and specific heat of water is 4200 J `kg^(-1)K^(-1).`

Explanation

The heat required to convert 10 g of ice at `0^(@)C` to water at `0^(@)C` is
`Q_(1)=m xx L`
Substitute `m=10 g=10^(-2)kg` and
`L=336xx10^(3)J kg^(-1)` in (1)
`Q_(1)=10^(-2)xx336xx10^(3)J`
`Q_(1)=3360 J" (1)"`
The heat required to raise the temperature of water from `0^(@)C" to "100^(@)C` is
`Q_(2)= mxx s xx Delta theta" (2)"`
Substituting `m=10^(-2) kg , and s=4200 J kg^(-1) K^(-1)`
`Delta theta = 100^(@)C,`
`Q_(2)=10^(-2)xx4200xx100`
`Q_(2)=4200 J" (3)"`
Substituting `m = 10^(-2) Kg//L =2260xx10^(3) J kg^(-1)`
`Q_(3)= m xx L =10^(-2)xx2260 xx 10^(3)`
`Q_(3)=22600 J" (4)"`
where `Q_(3)` is the heat required to convert water at `100^(@)C` to vapour
The total heat energy required is given by,
`Q=Q_(1)+Q_(2)+Q_(3)`
`Q=3360 J + 4200 J +22600 J`
`Q=30160 J`