A piece of ice of mass 60 g is dropped into 140 g of water at `50^(@)C`.
Calculate the final temperature of water when all the ice has melted.
(Assume no heat is lost to the surrounding)
Specific heat capacity of water = `4.2Jg^(-1)k^(-1)`
Specific latent heat of fusion of ice = `336Jg^(-1)`

Explanation

Let final temperature of water = `x""^(@)C`
`{:("Ice","Water"),(m_(1)=60g,m_(2)=140g),(T_(1)=0^(@)C,T_(1)=50^(@)C),(T_(2)=x^(@)C,T_(2)=x^(@)C),("Rise in temp.","fall in temp."),((DeltaT)=(x-0)^(@)C=x^(@)C,(DeltaT)=(50-x)^(@)C):}`
Heat gained by ice = `M_(1)L+m_(1)cDeltaT`
= `(60xx336+60xx4.2xxx)J`
Heat lost by water = `m_(2)cDeltaT`
= `140xx4.2xx(50-x)J`
Applying, the principle of mixtures,
`140xx4.2xx(50-x)=60xx336+60xx4.2xxx`
`rArr4.2(7000-140x-60x)=60xx336`
`rArr7000-200x=(60xx336)/4.2`
= `(60xx336)/42xx10`
= 4800
`rArr200x=2200`
`rArrx=11^(@)C`
`therefore` Final temperature of water = `11^(@)C`