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 cc = 0.39 J g-1 K-1
The specific heat capacity of water cw = 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,
Hence, final temperature = 47°C