The amount of heat needed to heat 200 grams of ice at $-10^{\circ} \mathrm{C}$ to convert it into water at…
The amount of heat needed to heat 200 grams of ice at $-10^{\circ} \mathrm{C}$ to convert it into water at $30^{\circ} \mathrm{C}$ is
Specific heat capacity of ice $=2100 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
Specific heat capacity of water $=4186 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$
Latent heat of fusion of ice $=3.35 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$
$96316 \mathrm{~J}$
$67000 \mathrm{~J}$
$92116 \mathrm{~J}$
$71200 \mathrm{~T}$
Solution
We have
$\Delta \mathrm{Q}=\mathrm{mS}_{\mathrm{i}}\left[-10^{\circ}-0^{\circ}\right]-\mathrm{mL}_{\mathrm{f}}+\mathrm{mS}_{\mathrm{w}}\left[0^{\circ}-30^{\circ}\right]$
$\begin{aligned} & =-\mathrm{mS}_{\mathrm{i}}\left(+10^{\circ}\right)-\mathrm{mL}_{\mathrm{f}}-\mathrm{mS}_{\mathrm{w}}\left(30^{\circ}\right) \\ & =-0.2 \times 2100 \times 10-0.2 \times 3.35 \times 10^5-0.2 \times 4186 \times 30 \\ & =-96316 \mathrm{~J}\end{aligned}$
-ve sign indicate that heat is absorbed
So, $\Delta Q_{\text {absorbed }}=96316 \mathrm{~J}$