200 g of hot water at 80°C is added to 400 g of cold water at 10°C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg^-1 K^-1.

Mass of hot water = 200 g

Temperature of hot water = 80° C

Mass of cold water = 400 g

Temperature of cold water = 10° C

Let final temperature be t

Fall in temperature of hot water = (80 – t)°C

Rise in temperature of cold water = (t - 10)°C

The specific heat capacity of water c_w = 4200 J kg^-1 K^-1 = 4.2 J g^-1 K^-1

Heat energy given by hot water = mc△t
= 200 x 4.2 x (80 – t)     [Equation 1]

Heat energy taken by cold water = 400 x 4.2 x (t - 10)     [Equation 2]

Assuming that there is no loss of heat energy,

Heat energy given by hot water = Heat energy taken by cold water

Equating equations 1 & 2, we get,

$200 \times 4.2 \times (80 - t) = 400 \times 4.2 \times (t - 10) \\[0.5em] 2 \times (80 - t) = 4\times (t - 10) \\[0.5em] 160 - 2t = 4t - 40 \\[0.5em] 160 + 40 = 4t + 2t \\[0.5em] \\[0.5em] 200 = 6t \\[0.5em] \Rightarrow t = \dfrac{200}{6} \\[0.5em] \Rightarrow t = 33.3°C$

Hence, final temperature = 33.3°C