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 (c) = 4.2 Jg^-1k^-1,
Specific latent heat of fusion of ice (L) = 336 Jg^-1

Let final temperature of water be T °C.

Heat energy taken by the ice to melt at 0 °C

Q₁ = mass of ice x specific latent heat of fusion of ice
= 60 x 336 = 20,160 J

Heat energy taken by the melted ice to raise it's temperature from 0 °C to T °C.

Q₂ = mass of ice x specific heat capacity of water x rise in temperature
= 60 x 4.2 x (T - 0) = 252T J

Total heat energy taken by ice = Q₁ + Q₂
= (20,160 + 252 T) J

Heat energy given by water in fall of it's temperature from 50 °C to T °C
= mass of water x specific heat capacity of water x fall in temperature
= 140 x 4.2 x (50 - T)
= 588 (50 - T) J

If there is no heat loss,

heat energy given by water = total heat energy taken by ice

Substituting the values we get,

$588 (50 - T) = 20,160 + 252 T \\[0.5em] (588 \times 50) - 588 T = 20,160 + 252 T \\[0.5em] 29,400 - 588 T = 20,160 + 252 T \\[0.5em] 29,400 - 20,160 = 588 T + 252 T \\[0.5em] 9240 = 840 T \\[0.5em] \Rightarrow T = \dfrac{9240}{840} \\[0.5em] \Rightarrow T = 11 °C$

Hence, T = 11 °C