Question

What will be the final temperature if 50g of water at exactly 0°C is added to 250g of water at 90°C?

Answer 1

Given:

$t_1=0^\circ\rm C$

$t_2=90^\circ\rm C$

$m_1=50\:\rm g$

$m_2=250\:\rm g$

The heat balance gives

$$Q_1=Q_2\\ cm_1(t_3-t_1)=cm_2(t_2-t_3)$$

Hence, the final temperature

$$t_3=\frac{m_1t_1+m_2t_2}{m_1+m_2}$$

$$t_3=\frac{50*0+250*90}{50+250}=75^\circ\rm C$$