A piece of metal weighing $100 \mathrm{~g}$ is heated to $80^{\circ} \mathrm{C}$ and dropped into $1…
A piece of metal weighing $100 \mathrm{~g}$ is heated to $80^{\circ} \mathrm{C}$ and dropped into $1 \mathrm{~kg}$ of cold water in an insulated container at $15^{\circ} \mathrm{C}$. If the final temperature of the water in the container is $15.69^{\circ} \mathrm{C}$, the specific heat of the metal in $\mathrm{J} / \mathrm{g}^{\circ} \cdot \mathrm{C}$ is:
$0.38$
$0.24$
$0.45$
$0.13$
Solution
Heat exchanged is given by, $\mathrm{Q}=\mathrm{m} \times \mathbf{s} \times\left(\mathrm{T}_2-\mathrm{T}_1\right)$ where $\mathbf{s}=$ specific heat
So, when heated metal is dropped into the cold water, we can write:
$\begin{aligned}
& \mathrm{m}_{\text {metal }} \times \mathrm{s}_{\text {metal }} \times\left(\mathrm{T}_2-\mathrm{T}_1\right)_{\text {metal }}=\mathrm{m}_{\text {water }} \times \mathrm{s}_{\text {water }} \times\left(\mathrm{T}_2-\mathrm{T}_1\right)_{\text {water }} \\
& 0.1 \times \mathrm{s}_{\text {metal }} \times(80-15.69)=1 \times 4.18 \times(15.69-15) \\
& \mathrm{s}_{\text {metal }}=0.45 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}
\end{aligned}$