A calorimeter contains 400 gm of water at temperature of `5^@C`. Then, 200 gm of water at a temperature of `+10^@C` and 400 gm of ice at temp. of `-60^@C` are added. Specific heat of ice `=0.5 ` cal/gm. At equilibrium

Answer

AB

Explanation

`Q_(reqd) =400xx 1/2 xx(0-(-60))`
=12000 cal.
`Q_(av)=400xx1xx(5-0)+200xx1xx(10-0)`
=4000 cal
`Q_("by freezing water")=600 xx 80 =48000 cal `
`rArr ` 100 gm water freezes
`rArr 500 gm` ice, 500 gm water.