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.

Options

A. `0.450 kg`
B. `1 kg`
C. `0.50 kg`
D. `1.5 kg`

Answer

A

Explanation

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`.