50 grams of ice at 0 ∘ C are put into 125 grams of water at 2 0 ∘ C in a calorimeter of negligible heat capacity. When the system has reached equilibrium, how much of the ice remains? (latent heat of

Question

50 grams of ice at 0 ∘ C are put into 125 grams of water at 2 0 ∘ C in a calorimeter of negligible heat capacity. When the system has reached equilibrium, how much of the ice remains? (latent heat of fusion of water L f ​ = 80 cal/g, specific heat of water c = 1 cal/g ∘ C ) .

TestHub · Accepted Answer

Heat given by water cooling to 0 ∘ C : Q = 125 × 1 × 20 = 2500 cal Ice melted by this heat: m = 80 2500 ​ = 31.25 g Ice melts only partly, since 31.25 g is less than 50 g . Ice left: 50 − 31.25 = 18.75 ≈ 19 g

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