A cube of ice of mass 30 g at `0^@`C is added into 200 g of water at `30^@`C. Calculate the final temperature of water when whole of the ice cube has melted.
Given: specific latent heat of ice = 80 cal `g^(-1)`, specific heat capacity of water = 1 cal `g^(-1)^@C^(-1)`

Explanation

Let final temperature of water be `t^@`C.
Heat energy taken by the ice to melt at `0^@`C
`Q_1` = mL
= 30 g `xx 80" cal "g^(-1)` = 2400 cal
Heat energy taken by the melted ice to raise its temperature from `0^@C to t^@`C
`Q_2` = mass `xx` specific heat capacity `xx` rise in temperature
= 30 g `xx` 1 cal `g^(-1)^@C^(-1) xx t^@` C
= 30 t cal
Total heat energy taken by ice = `Q_1 + Q_2`
= (2400 + 30 t) cal
Heat energy given by water in fall of its temperature from `30^@` C to `t^@`C
= mass `xx` specific heat capacity `xx` fall in temperature
= 200 g `xx` 1 cal `g^(-1)^@C^(-1) xx (30 – t)^@C`
= 200 (30 – t) cal. ....(ii)
If there is no heat loss,
heat energy given by water
= total heat energy taken by ice
200 (30 - 1) = 2400 + 30 t
or 6000 - 200 t = 2400 + 30 t
230 t = 3600
or t=`(3600)/(230)=15.65^@C`