Question

In an experiment on the specific heat of a metal a $0.20kg$ block of the metal at $150^{\circ }C$ is dropped in a copper calorimeter (of water equivalent $0.025kg$) containing $150c{m}^3$ of water at $27^{\circ }C$. The final temperature is $40^{\circ }C$. Compute the specific heat (in $Jk{g}^{-1}$ ${}^{\circ }{C}^{-1}$ of the metal.

Answer 1

Answer: 434.3

Let specific heat of the metal is $C$, the heat lost by metal block.
$=mC\Delta T$
$=0.20C(150-40)$
$=22C$
Mass of water in the calorimeter
$=dV=1000\times (150\times 10^{-6}=0.15kg$
Heat gained by water and calorimeter
$=m_{w}C\Delta T+C\Delta T$
$=(0.15\times 4200+0.025\times 4200)\times (40-27)$
$=735\times 13$
$=9555J$
By principle of calorimetry, we have
$22C=9555$
$\therefore C=\frac{9555}{22}$
$=434.3J/kg{-}^{\circ }C$