Question

200 grams metal at 1000C is placed on calorimeter which contains 300 grams water at 180C. If the specific heat of metal is 140 J/kg0C, determine the final temperature of the system!

Answer 1

Given:

$m_1=200\:\rm g$

$t_1=100^\circ\rm C$

$m_2=300\:\rm g$

$t_2=18^\circ\rm C$

$c_1=140\:\rm J/kg^\circ\rm C$

$c_2=4184\:\rm J/kg^\circ\rm C$

The heat balance equation says

$$c_1m_1(t_1-t_3)=c_2m_2(t_3-t_2)$$

Hence, the final temperature of system

$$t_3=\frac{c_1m_1t_1+c_2m_2t_2}{c_1m_1+c_2m_2}$$

$$t_3=\frac{140*0.2*100+4184*0.3*18}{140*0.2+4184*0.3}=20^\circ\rm C$$