Two calorimeters of water equivalent 25 g and 60 g are initially at `0^(@)C`. Some water at `50^(@)C` is poured in the first calorimeter and after thermal equilibrium is attained, the first calorimeter is emptied into the second one. If the final temperature of water and the second calorimeter becomes `25^(@)C`, find the mass of water.

Explanation

Let at equilibrium, temperature of the 1st calorimeter `=theta`, mass of water added = m.
Since, heat lost by hot water = heat gained by the 1st calorimeter
`therefore" "mxx1xx(50-theta)=25xx1xx theta`
`"or, "theta=(50m)/(25+m)" ...(1)"`
In the second calorimeter, heat lost by mg of water at `theta=mxx1xx(theta-25)` and heat gained by the calorimeter `=60xx25`
`therefore" "m(theta-25)=60xx25`
`"or, "theta=(60xx25)/(m)+25" ...(2)"`
Equating equations (1) and (2),
`(50m)/(25+m)+(60xx25)/(m)+25`
`"or, "(50m)/(25+m)=25((60)/(m)+1)" or, "(2m)/(25+m)=(60+m)/(m)`
`"or, "2m^(2)=(60+m)(25+m)`
`"or, "2m^(2)=1500+85m+m^(2)`
`"or, "m^(2)-85m-1500=0`
`"or "(m-100)(m+10)=0`
`therefore" "m=-15" or, "m=100`
Since negative mass is not acceptable, m = 100.
`therefore` The required mass of water is 100 g.