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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).` |
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Answer» 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` |
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