2 kg ice at −20∘C is mixed with 5 kg water at 20∘C. Then final amount of water in the mixture would beGiven specific heat of ice =0.5 cal/g∘C,Specific heat of water =1 cal/g∘C,Latent heat of fusion for ice =80 cal/g.

2 kg ice at −20∘C is mixed with 5 kg water at 20∘C. Then final a

1.	2 kg ice at −20∘C is mixed with 5 kg water at 20∘C. Then final amount of water in the mixture would beGiven specific heat of ice =0.5 cal/g∘C,Specific heat of water =1 cal/g∘C,Latent heat of fusion for ice =80 cal/g.
Answer» $2 kg$ ice at $- 20^{\circ} C$ is mixed with $5 kg$ water at $20^{\circ} C$ . Then final amount of water in the mixture would be Given specific heat of ice $= 0.5 {cal/g}^{\circ} C$ , Specific heat of water $= 1 {cal/g}^{\circ} C$ , Latent heat of fusion for ice $= 80 cal/g$ .

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2 kg ice at −20∘C is mixed with 5 kg water at 20∘C. Then final amount of water in the mixture would beGiven specific heat of ice =0.5 cal/g∘C,Specific heat of water =1 cal/g∘C,Latent heat of fusion for ice =80 cal/g.

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