Heat energy of $184\ \text{kJ}$ is given to ice of mass $600\ \text{g}$ at $-12^\circ\text{C}$. Specific heat of ice is $2222.3\ \text{J kg}^{-1}\,^\circ\text{C}^{-1}$ and latent heat of ice is $336\ \text{kJ kg}^{-1}$.
(A) Final temperature of system will be $0^\circ\text{C}$.
(B) Final temperature of the system will be greater than $0^\circ\text{C}$.
(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:
Answer: (A) A and D only
Heat to warm the ice to $0^\circ\text{C}$: $0.6\times2222.3\times12\approx16\ \text{kJ}$.
Remaining heat $184-16=168\ \text{kJ}$ melts $\dfrac{168}{336}=0.5\ \text{kg}$ of ice.
So $0.1\ \text{kg}$ ice and $0.5\ \text{kg}$ water remain at $0^\circ\text{C}$: ice : water $=1:5$. A and D are correct.
Solution by Sreeraj P, M.Sc Physics