1.

`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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