A vessel of mass 100 g contains 150 g of water at `30^@`C. How much ice is needed to cool it to `5^@`C ? Take specific heat capacity of material of vessel = 0-4 J `g^(-1) K^(-1)`, specific latent heat of fusion of ice = 336 J g- and specific heat capacity of water = 4.2 J `g^(-1) K^(-1)`.

Explanation

Heat energy imparted by vessel and water contained in it in cooling from `30^@`C to `5^@`C, is used in melting ice and then raising the temperature of melted ice from `0^@`C to `5^@`C.
Heat energy imparted by vessel
`Q_1 = 100 xx 0.4 (30 - 5) = 1000` J ...(i)
Heat energy imparted by water
`Q_2 = 150 xx 4.2 xx(30 - 5) = 15750` J ....(ii)
Let m g of ice be used, then
Heat energy taken by ice to melt
`Q_3 = m xx` 336 J ....(iii)
and heat energy taken by the melted ice to raise its temperature from `0^@`C to `5^@`C
`Q_4 = m xx 4.2 xx (5 - 0)` = 21 m J ....(iv)
By law of conservation of energy,
heat energy imparted by vessel and water
= heat energy taken by ice and melted ice
or `Q_1+Q_2 = Q_3 +Q_4`
i.e., 1000 + 15750 = 336 m + 21 m
or 357 m = 16750
or m=`(16750)/(357)` = 46.92 g
Thus, 46.92 g of ice is used.