Question

A 50-g piece of metal at 95°C is dropped into 250g of water at 17°C and warms it to 19.4°C. What is the specific heat of the metal?

Answer 1

Given:

$t_1=95^\circ\rm C$

$t_2=17^\circ\rm C$

$t_3=19.4^\circ\rm C$

$m_1=50\:\rm g$

$m_2=250\:\rm g$

The heat balance gives

$$Q_1=Q_2\\ cm_1(t_1-t_3)=c_wm_2(t_3-t_2)$$

Hence, the specific heat of the metal

$$c=c_w\frac{m_2(t_3-t_2)}{m_1(t_1-t_3)}$$

$$c=\rm4.186\: J/g*\frac{250(19.4-17)}{50(95-19.4)}=0.664\: J/g$$