Analysis:
heat loss by \(50 kg\) of hot water at \(80^{\circ} C\)
= heat gain by \(7.0 kg\) of water at \(20^{\circ} C\)
let the final temperature be \(\theta\)
\(m c \Delta \theta=m c \Delta \theta\)
\(50 \times c \times(80 \theta)=70 \times c \times(\theta-20)\)
\(5(80-\theta)=7(\theta-20)\)
\(400-5 \theta=7 \theta-140\)
\(400+140=7 \theta+5 \theta\)
\(\frac{540}{12}=\frac{12 \theta}{12}, \theta=45^{\circ} C\)