Find the temperature of the water after passing 5 g of steam at 100°C through 500 g of water at 10°C.         (16.2°C)
(Note: Specific heat of water is $4200 J〖kg〗^{-1} K^{-1}$, Latent heat of vaporization of water is $2.26 \times 〖10〗^{6} Jkg^{-1})$.
Difficulty: Hard
Solution:    Mass of stream = $m_{1}= 5 g$ =$\frac{5}{1000}$ kg = 0.005 kg
Temperature of stream = $T_{1}$= 100°C
Mass of water = $m_{2}$ = 0.5 kg
Temperature of water = $T_{2}$= 10°C
Final temperature = $T_{3}$= ?

Case I:
Latent heat lost by stream = Q1 = mL
Q1 = $0.005\times 2.26 \times 10^{6}$ = $11.3 \times 10^{3}$ = 11300 J

Case II:
Heat lost by stream to attain final temperature Q2 = $m_{1} c \triangle T$
Q2 = $0.005\times 4200 \times (100- T_{3})$
Q2 = $21(100- T_{3})$

Case III:
Heat gained by water Q3 = $m_{2} c \triangle T$
Q3 = $0.5\times 4200\times (T_{3} - 10)$
Q3 = $2100(T_{3}- 10)$
According to the law of heat exchange.
Heat lost by stream = heat gained by water
Q1 + Q2 = Q3
$11300 + 21 (100 – T_{3})$ = $2100 (T_{3}- 10)$
$11300 + 2100 - 21T_{3}$ = $2100 T_{3}$ – 21000
$13400 + 21000 – 21 T_{3}$ = $2100 T_{3} – 21 T_{3}$
34400 = $2121 T_{3}$
$T_{3}$= $\frac{34400}{2121}$
$T_{3}$= 16.2 °C