100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1 ∘C−1) at 80∘C. The final temperature of the mixture will be:

100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g

1.	100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1 ∘C−1) at 80∘C. The final temperature of the mixture will be:
Answer» $100 g$ of ice $(latent heat 80 {cal g}^{- 1})$ at $0^{\circ} C$ is mixed with $100 g$ of water (specific heat $1 {cal g}^{- 1 \circ} C^{- 1}$ ) at $80^{\circ} C$ . The final temperature of the mixture will be:

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100 g of ice (latent heat 80 cal g−1) at 0∘C is mixed with 100 g of water (specific heat 1 cal g−1 ∘C−1) at 80∘C. The final temperature of the mixture will be:

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