Question

$100g$ of ice at $-{10}^{\circ }C$ is heated on a gas stove till it forms water at ${80}^{\circ }C$. Calculate :

(a) Heat energy required to bring the ice to $0^{\circ }C$.

(b) The heat energy required to melt the ice

(c) Heat energy required to bring water to ${80}^{\circ }C$. [sp. heat capacity of ice $2J{g^{-1}}^{\circ }{C}^{-1}$; sp. heat capacity of water $4.2J{g}^{-1}{C}^{-1}$ and sp. latent heat of fusion of ice is $336J{g^{-1}}^{\circ }{C}^{-1}$.

Answer 1

Step 1: Given data

1. The quantity of ice is $100g$ and takes specific latent heat of fusion of ice $336J{g}^{-1}{C}^{-1}$,
2. Heat capacity of water $4.2J{g^{-1}}^{\circ }{C}^{-1}$
3. Specific heat capacity of ice $2J{g^{-1}}^{\circ }{C}^{-1}$

Step 2: Calculating (a) heat energy required to bring ice at $0^{\circ }C$

Heat energy required to bring ice at $0^{\circ }C$

$$\begin{gathered} =mc{\theta }_R \\ =100\times 2\times (0-(-10\left)\right) \\ =100\times 2\times 10 \\ =2000J \end{gathered}$$

Step 3: Calculating the (b) heat energy required to melt the ice

Heat energy required to melt ice

$$\begin{gathered} =mL \\ =100\times 336 \\ =33600J \end{gathered}$$

Step 4: Calculating the (c) heat energy required to bring water to the temperature of ${80}^{\circ }C$

Heat energy required to bring water to ${80}^{\circ }C$

$$\begin{gathered} =mc{\theta }_R \\ =100\mathrm{g}\times 4.2{\mathrm{Jg}}^{-10}{\mathrm{C}}^{-1}\times 80°\mathrm{C} \\ =33600\mathrm{J}. \end{gathered}$$

Therefore,

(a) The heat energy required to bring ice at $0^{\circ }C$ is $2000J.$

(b) The heat energy required to melt the ice is $33600J.$

(c) The heat energy required to bring water to the temperature of ${80}^{\circ }C$ is $33600J.$