Question

75 g of water at ${10}^{0}C$ is heated by supplying $25200J$ of heat energy. If the specific heat of water is $4.2J{g}^{-1}$ ${}^0{C}^{-1}$. Calculate the final temperature of water.

Answer 1

Given:

mass of water, $m=75g$

Heat supplied, $Q=25200J$

Specif heat, $C=4.2J{g}^{-1}{C}^{-1}$

intital temperature, $T_1=10°C$,

we know,

$Q=mC\Delta T$

$⟹\Delta T=\frac{Q}{mC}=\frac{25200}{75\times 4.2}=80°C$

$\Delta T=T_2-T_1⟹T_2=\Delta T+T_2=80+10=90°C$

The final temperature of water is 90°C.