Question

A 20 g sample of ice at -10 °C is mixed with 100 g water at 80 °C. Calculate the final temperature of the mixture.

Answer 1

Let the final temperature be T

Heat released by water = mass $\times$ specific heat capacity $\times$ temperature change

The specific heat of liquid water is 4.18 J/g°C

$= 100 \times 4.18 \times (80 - T)$

Heat absorbed by ice on melting = mass $\times$ heat of fusion

The heat of fusion of ice is 335 J/g

$= 20 \times 335 = 6700 \;J$

Heat absorbed by melted ice = mass $\times$ specific heat capacity $\times$ temperature change

$= 20 \times 4.18 \times (T - (-10)) = 83.6(T+10)$

Total heat released = total heat absorbed

$100 \times 4.18 \times (80 - T) = 6700 + 83.6(T+10) \\ 33440 -418T = 6700 +83.6T + 836 \\ 25904 = 501.6T \\ T = \frac{25904}{501.6} \\ T = 51.64 \; °C$

Answer: 51.64 °C