Question

Heat energy of $184 \,kJ$ is given to ice of mass $600\, g$ at $-12^{\circ} C$, Specific heat of ice is $2222.3 \,J\, kg ^{-1^{\circ}} C ^{-1}$ and latent heat of ice in $336 \,kJ / kg ^{-1}$
(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 =0.600 \,kg \, \text { at }-12^{\circ} C $
$ S =222.3 \, J / kg /{ }^{\circ} C $
$ L =336 \times 10^3 J / kg $
$ Q _1=0.600 \times 2222.3 \times 12=16000.56 \, J$
Remaining heat $\Delta Q_1=184000-16000.56$
$=167999.44 \, J$
For meeting at $0^{\circ} C$
$\Delta Q_2=0.600 \times 336000=201600 J$ needed
$\therefore 100 \%$ ice is not melted
Amount of ice melted
$167999.44= m \times 336000=0.4999 \,kg$
$\therefore$ mass of water $=0.4999 \, kg$
Mass of ice $=0.1001$
$\therefore \text { Ratio }=\frac{0.1001}{0.4999} \approx 1: 5$