A $50 \mathrm{~g}$ ice cube at $-10^{\circ} \mathrm{C}$ is added to $200 \mathrm{~g}$ of water at…
A $50 \mathrm{~g}$ ice cube at $-10^{\circ} \mathrm{C}$ is added to $200 \mathrm{~g}$ of water at $30^{\circ} \mathrm{C}$. The final temperature of the mixture is (specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}$, latent heat of fusion of ice $=80 \mathrm{cal} \mathrm{g}^{-1}$ specific heat of ice $=0.5 \mathrm{cal} \mathrm{g}^{-1}$ ${ }^{\circ} \mathrm{C}^{-1}$ )
$20{ }^{\circ} \mathrm{C}$
$7^{\circ} \mathrm{C}$
$12{ }^{\circ} \mathrm{C}$
$10^{\circ} \mathrm{C}$
Solution
Mass of ice, $\mathrm{m}=50 \mathrm{~g}$
Temperature, $\mathrm{T}=-10^{\circ} \mathrm{C}$
Added mass, $\mathrm{m}^{\prime}=200 \mathrm{~g}$
Temperature at which mass added, $\mathrm{T}^{\prime}=30^{\circ} \mathrm{C}$ final temperature of mixture, $\mathrm{T}_{\mathrm{f}}=$ ?
Heat lost $=$ heat gained
$
\begin{aligned}
& \mathrm{m}^{\prime} \mathrm{C}_{\mathrm{w}} \Delta \mathrm{T}_{\mathrm{w}}=\mathrm{mC}_{\mathrm{i}} \Delta \mathrm{T}_{\mathrm{i}}+\mathrm{mL}^{\mathrm{L}} \\
& 200 \times 1 \times\left(30-\mathrm{T}_{\mathrm{f}}\right)=50 \times 0.5 \times\left(\mathrm{T}_{\mathrm{f}}+10\right)+50 \times 80 \\
& \mathrm{~T}_{\mathrm{f}}=7^{\circ} \mathrm{C}
\end{aligned}
$