Thermodynamics Question 12
Question 12 - 29 January - Shift 2
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:
(1) A and D only
(2) B and D only
(2) A and E only
(4) A and C only
Show Answer
Answer: (1)
Solution:
Formula: Specific heat
$\Delta Q=184 \times 10^{3}$
$m=0.600 kg$ 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 $=0.1001$
$\therefore$ Ratio $=\frac{0.1001}{0.4999} \approx 1: 5$
