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`10 gm` of ice at `-20^(@)C` is dropped into a calorimeter containing `10 gm` of water at `10^(@)C`, the specific heat of water is twice that of ice. When equilibrium is reached the calorimeter will contain:A. `20 gm` of waterB. `20 gm` of iceC. `10 gm` ice and `10 gm` of waterD. `5 gm` ice and `15 gm` of water |
Answer» Correct Answer - C `Q_(1) = 10xx1xx10=100cal` `Q_(2) = 10xx0.5(0-(-20))+10xx80` `=(100+800)cal = 900cal` As `Q_(1) lt Q_(2)`, so ice will not completely melt and final temeprature = `0^(@)C` As heat given by water in cooling up to `0^(@)C` is only just sufficient to increase the temperature of ice from `-20^(@)C` to `0^(@)C` , hence mixture in equilibrium will consist of `10 gm` of ice and `10 gm` of water, both at `0^(@)C`. |
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