45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18°C is added to it. The contents are stirred till a final constant temperature is reached. Calculate the final temperature. The specific heat capacity of copper is 0.39 J g^-1 K^-1 and that of water is 4.2 J g^-1 K^-1. State the assumptions used.

Given,

Mass of water = 45 g

Let the final constant temperature reached be t°C

Fall in temperature of water = (50 – t)°C

Mass of copper = 50 g

Rise in temperature of copper = (t - 18)°C

The specific heat capacity of the copper c_c = 0.39 J g^-1 K^-1

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

Heat energy given by water = mc△t
= 45 x 4.2 x (50 – t)     [Equation 1]

Heat energy taken by copper = 50 x 0.39 x (t - 18)     [Equation 2]

Assuming that there is no loss of heat energy

Heat energy given by water = Heat energy taken by copper

Equating equations 1 & 2, we get,

$45 \times 4.2 \times (50 - t) = 50 \times 0.39 \times (t - 18) \\[0.5em] \Rightarrow 189 \times (50 - t) = 19.5 \times (t - 18) \\[0.5em] \Rightarrow (189 \times 50) - (189 \times t) = (19.5 \times t) - (19.5 \times 18) \\[0.5em] \Rightarrow (9450) - (189 \times t)] = (19.5 \times t) - (351) \\[0.5em] \Rightarrow 9450 + 351 = (19.5 t) + (189 t) \\[0.5em] \Rightarrow 9801 = (208.5t) \\[0.5em] \Rightarrow t = \dfrac{9801}{208.5} \\[0.5em] \Rightarrow t = 47.0072°C \approx 47°C$

Hence, final temperature = 47°C