A lump of 0.10kg of ice at −10∘C is put in 0.15kg of water at 20∘C. How much water and ice will be found in the mixture when it has reached in thermal equilibrium. (Specific heat of water =1kcalkg−1, specific heat of ice =0.50kcalkg−1 while its latent heat =80kcalkg−1)

A lump of 0.10kg of ice at −10∘C is put in 0.15kg of water at 20�

1.	A lump of 0.10kg of ice at −10∘C is put in 0.15kg of water at 20∘C. How much water and ice will be found in the mixture when it has reached in thermal equilibrium. (Specific heat of water =1kcalkg−1, specific heat of ice =0.50kcalkg−1 while its latent heat =80kcalkg−1)
Answer» A lump of $0.10 k g$ of ice at $- 10^{\circ} C$ is put in $0.15 k g$ of water at $20^{\circ} C$ . How much water and ice will be found in the mixture when it has reached in thermal equilibrium. (Specific heat of water $= 1 k c a l k g^{- 1}$ , specific heat of ice $= 0.50 k c a l k g^{- 1}$ while its latent heat $= 80 k c a l k g^{- 1}$ )

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A lump of 0.10kg of ice at −10∘C is put in 0.15kg of water at 20∘C. How much water and ice will be found in the mixture when it has reached in thermal equilibrium. (Specific heat of water =1kcalkg−1, specific heat of ice =0.50kcalkg−1 while its latent heat =80kcalkg−1)

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