$$2kg$$ of ice at −$$20$$∘C is mixed with $$5kg$$ of water at $$20o$$ C . The water content of final mixture is (Latent$$ heat of ice $$=$$ $$80$$ $$kcal/kg$$, specific heat of water $$=1k$$ cal kg−$$1$$∘C−$$1specific$$ heat of ice $$=0.5k$$ cal kg−$$1$$∘C−$$1)

Question

$$2kg$$ of ice at −$$20$$∘C is mixed with $$5kg$$ of water at $$20o$$ C . The water content of final mixture is (Latent$$ heat of ice $$=$$ $$80$$ $$kcal/kg$$, specific heat of water $$=1k$$ cal kg−$$1$$∘C−$$1specific$$ heat of ice $$=0.5k$$ cal kg−$$1$$∘C−$$1)

Infinity Learn · Accepted Answer

Heat lost by $$water=ms$$ θ, heat gained by ice $$=MS$$ θ$$+M$$'L maximum heat loss by $$water=5$$×$$20$$×$$1kcal=100kcal$$ heat gained by ice to reach $$00C=2$$×$$20$$×$$0.5=20$$ kcal  remaining  heat $$100-20=80Kcal$$ it is just enough to melt $$1$$ kg of ice. so total water in the $$mixture=5+1=6kg$$

2kg of ice at $- 20^{\circ} C$ is mixed with 5kg of water at 20^o C . The water content of final mixture is (Latent heat of ice = 80 kcal/kg, specific heat of water $= 1 k cal {kg}^{- 1}^{\circ} C^{- 1}$ specific heat of ice $= 0 .5 k cal {kg}^{- 1}^{\circ} C^{- 1}$ )

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2kg of ice at −20∘C is mixed with 5kg of water at 20o C . The water content of final mixture is (Latent heat of ice = 80 kcal/kg, specific heat of water =1k cal kg−1∘C−1specific heat of ice =0.5k cal kg−1∘C−1)

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2kg of ice at $- 20^{\circ} C$ is mixed with 5kg of water at 20^o C . The water content of final mixture is (Latent heat of ice = 80 kcal/kg, specific heat of water $= 1 k cal {kg}^{- 1}^{\circ} C^{- 1}$ specific heat of ice $= 0 .5 k cal {kg}^{- 1}^{\circ} C^{- 1}$ )