$50 g$ of ice at $0^{\circ} C$ is dropped into $200 g$ of water at $4 0^{\circ} C$ in an insulated container (negligible heat capacity). Take the heat of fusion of ice as $334 J g^{- 1}$ and the specific heat of water as $4.18 J g^{- 1} K^{- 1}$ . Find the final temperature in $^{\circ} C$ (one decimal).

Numerical answer type, answer in degC, tolerance 0.1.

Answer: 16 degC

Heat available from cooling the warm water to

0^{\circ} C

200 \times 4.18 \times 40 = 33440 J

, more than the

50 \times 334 = 16700 J

needed to melt all the ice, so the ice melts completely. The remaining

16740 J

warms all

250 g

T_{f} = 16740/ (250 \times 4.18) = 16.0 2^{\circ} C

, i.e. 16.0.