Ice at $-20^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of water at $40^{\circ} \mathrm{C}$, When the…
Ice at $-20^{\circ} \mathrm{C}$ is added to $50 \mathrm{~g}$ of water at $40^{\circ} \mathrm{C}$, When the temperature of the mixture reaches $0^{\circ} \mathrm{C},$ it is found that 20 $\mathrm{g}$ of ice is still unmelted. The amount of ice added to the
water was close to (Specific heat of water $=4.2 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Specific heat of Ice $=2.1 \mathrm{~J} / \mathrm{g} /{ }^{\circ} \mathrm{C}$
Heat of fusion of water at $\left.0^{\circ} \mathrm{C}=334 \mathrm{~J} / \mathrm{g}\right)$
$50 \mathrm{~g}$
$100 \mathrm{~g}$
$60 \mathrm{~g}$
$40 \mathrm{~g}$
Solution
Let m gram of ice is added. From principal of calorimeter heat gained (by ice) $=$ heat lost (by water)
$\begin{array}{l}
\therefore 20 \times 2.1 \times \mathrm{m}+(\mathrm{m}-20) \times 334 \\
=50 \times 4.2 \times 40 \\
376 \mathrm{~m}=8400+6680 \\
\mathrm{~m}=40.1
\end{array}$