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