Question

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

• A. 7 kg
• B. 6 kg
• C. 4 kg
• D. 2 kg

Answer 1

Answer: (B)

6 kg

Answer 2

Heat released by 5 kg of water when its temperature falls from 20$^{\circ}$C to 0$^{\circ}$C is
$Q_1 =mc \Delta \theta =(5)(10^3)(20-0)=10^5 cal$

when 2 kg ice at -20$^{\circ}$C comes to a temperature of 0$^{\circ}$C, it takes an energy
$Q_2 =mc\Delta \theta =(2)(500)(20) =0.2 \times 10^{5}cal$

The remaining heat $Q=Q_1-Q_2 =0.8 \times 1^5$ cal will melt a mass m of the ice, where
$\, \, \, \, \, \, \, \, \, \, \, m=\frac{Q}{L} =\frac{0.8 \times 10^5}{80 \times10^3}=1kg$
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.