Water of mass $m_1 = 2$ kg at a temperature of $t_1 = 80°$C is mixed with water of mass $m_2 = 3$ kg at a temperature of $t_2 = 20°$C. Find the temperature of the mixture. Neglect heat loss to the surroundings.
The hot water gives up as much heat as the cold water absorbs (the specific heat capacity is the same and cancels out):
$$c m_1 (t_1 - t) = c m_2 (t - t_2)$$
$$t = \frac{m_1 t_1 + m_2 t_2}{m_1 + m_2} = \frac{2 \cdot 80 + 3 \cdot 20}{2 + 3} = \frac{160 + 60}{5} = 44°\text{C}$$
$t = 44°$C