Heat energy of $$184$$ kJ is given to ice of mass $$600$$ g at $$-12°$$C, Specific heat of ice is $$2222.3$$ J kg$$^{-1}$$ °C$$^{-1}$$ and latent heat of ice is $$336$$ kJ kg$$^{-1}$$.
(A) Final temperature of system will be $$0°$$C
(B) Final temperature of the system will be greater than $$0°$$C
(C) The final system will have a mixture of ice and water in the ratio of $$5 : 1$$
(D) The final system will have a mixture of ice and water in the ratio of $$1 : 5$$
(E) The final system will have water only
Choose the correct answer from the options given below:

Solution :

Given :

Mass of ice,

$$m = 600\text{ g} = 0.6\text{ kg}$$

Initial temperature,

$$-12^\circ\text{C}$$

Heat supplied,

$$Q = 184\text{ kJ}$$

Specific heat of ice,

$$s = 2222.3\text{ J kg}^{-1}\ ^\circ\text{C}^{-1}$$

Latent heat of fusion,

$$L = 336\text{ kJ kg}^{-1}$$

Heat required to raise temperature of ice from $$-12^\circ\text{C}$$ to $$0^\circ\text{C}$$ :

$$Q_1 = ms\Delta T$$

$$= 0.6 \times 2222.3 \times 12$$

$$\approx 16000\text{ J}$$

$$= 16\text{ kJ}$$

Remaining heat :

$$Q_2 = 184 - 16$$

$$= 168\text{ kJ}$$

Heat required to completely melt the ice :

$$Q_m = mL$$

$$= 0.6 \times 336$$

$$= 201.6\text{ kJ}$$

Since available heat $$168\text{ kJ}$$ is less than $$201.6\text{ kJ}$$, all the ice will not melt.

Mass of ice melted :

$$m' = \frac{168}{336}$$

$$= 0.5\text{ kg}$$

Remaining ice :

$$0.6 - 0.5 = 0.1\text{ kg}$$

Water formed :

$$0.5\text{ kg}$$

Therefore,

Ice : Water

$$= 0.1 : 0.5$$

$$= 1 : 5$$

Also, final temperature remains $$0^\circ\text{C}$$.

Hence,

(A) Correct

(D) Correct

Final Answer :

A and D