Question

Heat energy of $184kJ$ is given to ice of mass $600g$ at $-12^{\circ }C$, Specific heat of ice is $2222.3Jk{g}^{-1^{\circ }}{C}^{-1}$ and latent heat of ice in $336kJ/k{g}^{-1}$
(A) Final temperature of system will be $0^{\circ }C$.
(B) Final temperature of the system will be greater than $0^{\circ }C$.
(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:

• A. A and D only
• B. B and D only
• C. A and E only
• D. A and C only

Answer 1

Answer: (A)

$\Delta Q=184\times 10^3$
$m=0.600kg\text{ at }-12^{\circ }C$
$S=222.3J/kg/{}^{\circ }C$
$L=336\times 10^{3}J/kg$
$Q_1=0.600\times 2222.3\times 12=16000.56J$
Remaining heat $\Delta Q_1=184000-16000.56$
$=167999.44J$
For meeting at $0^{\circ }C$
$\Delta Q_2=0.600\times 336000=201600J$ needed
$\therefore 100\%$ ice is not melted
Amount of ice melted
$167999.44=m\times 336000=0.4999kg$
$\therefore$ mass of water $=0.4999kg$
Mass of ice $=0.1001$
$\therefore \text{ Ratio }=\frac{0.1001}{0.4999}\approx 1:5$