$$2$$ kg of ice at $$-20$$°C is mixed with $$5$$ kg of water at $$20$$°C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water in the vessel in kg. It is given that the specific heat of water and ice are $$1$$ $$kcal/kg/$$°C and $$0.5$$ $$kcal/kg/$$°C respectively and the latent heat of fusion of ice is $$80$$ $$kcal/kg$$.

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Infinity Learn · Accepted Answer

Let m kg be the mass of ice melted into water. Heat lost by $$5$$ kg of water $$=$$ $$5$$ kg × $$1$$ $$kcal/kg/$$°C × $$20$$°C $$=$$ $$100$$ kcal. Heat gained $$=$$ $$mL+ms$$Δ$$T=m$$ kg × $$80$$ $$kcal/kg$$ $$+$$ $$2$$ kg × $$0.5$$ $$kcal/kg/$$°C × $$20$$°C $$=$$ $$80$$ m kcal $$+$$ $$20$$ kcal. Now, heat gained $$=$$ heat lost. Therefore, $$80$$ m $$+$$ $$20$$ $$=$$ $$100or$$ $$m=1$$ kg. Therefore, final mass of water $$=$$ $$5$$ kg $$+$$ $$1$$ kg $$=$$ $$6$$ kg.

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