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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:
$A$ and $D$ only
$B$ and $D$ only
$A$ and $E$ only
$A$ and $C$ only
Solution
$\Delta Q =184 \times 10^3$
$m =0.600\,kg \text { at }-12^{\circ}\,C$
$S =222.3\,J / kg /{ }^{\circ}\,C$
$L =336 \times 10^3 \,J / kg$
$Q _1=0.600 \times 2222.3 \times 12=16000.56\,J$
Remaining heat $\Delta Q _1=184000-16000.56$
$=167999.44\,J$
For meeting at $0^{\circ}\,C$
$\Delta Q _2=0.600 \times 336000=201600\,J \text { needed }$
$\therefore 100 \%$ 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 $=\frac{0.1001}{0.4999} \approx 1: 5$