InterviewSolution
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InterviewSolution
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When `0.400 kg` of water at `30^(@)C` is mixed with `0.150 kg` of water at `25^(@)C` contained in a calorimeter, the final temperature is found to be `27^(@)C`, find the water equivalent of the calorimeter.A. `0.450 kg`B. `1 kg`C. `0.50 kg`D. `1.5 kg` |
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Answer» Correct Answer - A Water of mass `m_(1)=0.150 kg` is taken in the calorimeter at temperature, `T_(1) = 25^(@)C` is mixed with `N` another known mass of pure water `m_(2) =0.400 kg` at a temperature `T_(2) = 30^(@)C` and final temperature is found to be `T=27^(@)C`. Let `s_(w)` and `W` be the heat capacity of water and the water equivalent of the calorimeter. Heat gained by (water+calorimeter) at temperature `T_(1)`= Heat lost by water at temperature `T_(2)` i.e., `m_(1)s_(w)(T-T_(1))+Ws_(w) (T-T_(1)) = m_(2)s_(w)(T_(2)-T)` or, `W = m_(2)((T_(2)-T)/(T-T_(1)))-m_(1)` Hence, `W = 0.400kg[(30^(@)C-27^(@)C)/(27^(@)C-25^(@)C)]-0.150 kg` `=0.450 kg`. |
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