Heat and Thermodynamics 1 Question 14
14. $2 kg$ of ice at $-20^{\circ} C$ is mixed with $5 kg$ of water at $20^{\circ} C$ in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are $1 kcal / kg /{ }^{\circ} C$ and $0.5 kcal / kg /{ }^{\circ} C$ while the latent heat of fusion of ice is $80 kcal / kg$
(2003, 2M)
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Solution:
- Heat released by $5 kg$ of water when its temperature falls from $20^{\circ} C$ to $0^{\circ} C$ is,
$$ Q _1=m c \Delta \theta=(5)\left(10^{3}\right)(20-0)=10^{5} cal $$
when $2 kg$ ice at $-20^{\circ} C$ comes to a temperature of $0^{\circ} C$, it takes an energy
$$ Q _2=m c \Delta \theta=(2)(500)(20)=0.2 \times 10^{5} cal $$
The remaining heat $Q=Q _1-Q _2=0.8 \times 10^{5}$ cal will melt a mass $m$ of the ice, where,
$$ m=\frac{Q}{L}=\frac{0.8 \times 10^{5}}{80 \times 10^{3}}=1 kg $$
So, the temperature of the mixture will be $0^{\circ} C$, mass of water in it is $5+1=6 kg$ and mass of ice is $2-1=1 kg$.
