10 gm ice at − 4 ∘ C is mixed with 41 gm of water at 2 0 ∘ C in an isolated container. ( S w a t er = 1 cal/gm ∘ C , S i ce = 1/2 cal/gm ∘ C , L f = 80 cal/gm), then choose the correct answers

Question

10 gm ice at − 4 ∘ C is mixed with 41 gm of water at 2 0 ∘ C in an isolated container. ( S w a t er ​ = 1 cal/gm ∘ C , S i ce ​ = 1/2 cal/gm ∘ C , L f ​ = 80 cal/gm), then choose the correct answers

TestHub · Accepted Answer

Heat needed by the ice to warm to 0 ∘ C and melt completely: Q i ce ​ = 10 × 2 1 ​ × 4 + 10 × 80 = 820 cal Heat given by the water cooling to 0 ∘ C : Q w a t er ​ = 41 × 1 × 20 = 820 cal The two are equal. So all the ice just melts and the temperature is 0 ∘ C . Water at the end: 41 + 10 = 51 gm

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