Question

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

Answer 1

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