The final temperature of the bath water is [tex]30.4^{\circ}[/tex]
Solution:
From given,
Mass of the hot water = m = 3 kg
Mass of cold water = M = 27 kg
Initial temperature of the hot water = [tex]T_{1h} = 79^{\circ}[/tex]
Initial temperature of the cold water = [tex]T_{1c} = 25^{\circ}[/tex]
The heat lost by the hot water will be equal to the heat gain by the cold water
Therefore,
[tex]mC(T_{1h} - T) = MC(T - T_{1c})[/tex]
[tex]3 \times 4.18 (79-T) = 27 \times 4.18(T - 25)\\\\12.54(79-T) = 112.86(T - 25)\\\\990.66 - 12.54T = 112.86T - 2821.5\\\\112.86T + 12.54T = 990.66 + 2821.5\\\\125.4T = 3812.16\\\\Divide\ both\ sides\ by\ 125.4\\\\T = 30.4[/tex]
Thus the final temperature of the bath water is [tex]30.4^{\circ}[/tex]
