When a calorimeter contains `40g` of water at `50^(@)C`, then the temperature falls to `45^(@)C` in `10` minutes. The same calorimeter contains `100g` of water at `50^(@)C`, it takes `20` minutes for the temperature to become `45^(@)C`. Find the water equivalent of the calorimeter.

Explanation

`(m_(t)s_(t) + W)/(t_(1)) = (m_(2)s_(2) + W)/(t_(2))` where `W` is the water equivalent
` implies (40xx1+W)/(10) = (100xx1+W)/(20)`
` implies 80+2W= 100+W implies W=20g`