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Calcuate the heat required to convert `0.6` kg of ice at - `20^(@)C`, kept in a calorimeter to steam at `100^(@)C` at atmospheric pressure. Given the specific heat capacity of ice`=2100 j kg^(-1) K^(-1)`, specific heat capacity of water `=4186 j kg^(-1) K^(-1)` latent heat ice `=3.35xx10^(5) j kg^(-1)` and latent heat of steam `=2.256xx10^(6)j kg^(-1)` |
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Answer» Correct Answer - `[1.8xx10^(6)j]` Here, `m=0.6 kg, T_(1)=20^(@)C T_(2)=100^(@)C` `s_(i)2100 j kg^(-1) K^(-1), K^(-1), s_(w)=4186 j kg ^(-1) K^(-1)`, `L_(i)=3.35xx10^(5) j kg ^(-1), L_(s)=2.256xx10^(6) j kg ^(-1)`, Heat required to convert ice at `-20^(@)C` to ice at `0^(@)C` `Q_(1)=ms_(i)Delta T_(1)=0.6xx3.35xx10^(5)=2.01xx10^(5)j =201000 j` Heat required to convert water at `0^(@)C` to `100^(@)C` `Q_(3)=ms_(w)Delta T_(2)=0.6xx4186xx(100-0)=251160 j` Heat required to convert water at `100^(@)C` into steam at `100^(@)C` `Q_(4)mL_(s)=0.6xx2.256xx10^(6)=1353600 j` Total heat required to convert `0.6` kg of ice at `-2^(@)C` into steam at `100^(@)C` is `Q=Q_(1)+Q_(2)+Q_(3)+Q_(4)` `=25200+201000+251160+1353600` `=1830960 j =1.83xx10^(6)j` |
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