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1 kg of ice at 0^(@)C is mixed with 1 kg...

1 kg of ice at `0^(@)C` is mixed with 1 kg of steam at `100^(@)C`. Find the equilibrium temperature and the final composition of the mixture. Given that latent heat of fusion of ice is `3.36xx10^(5)` J/kg and latent heat of vaporization of water is `2.27xx10^(6)` J/kg. Specific heat of water is `4200J//kg.^(@)C^(-1)`

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The correct Answer is:
`100^(@)C`, mixture =1335 g water +665 g steam
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1kg ice at -10^(@)C is mixed with 1kg water at 100^(@)C . Then final equilirium temperature and mixture content.

1kg ice at -10^(@) is mixed with 1kg water at 50^(@)C . The final equilibrium temperature and mixture content.

1kg ice at 0^(@)C is mixed with 1kg of steam at 100^(@)C what will be the composition of the system when thermal equilibrium is reached ? Latent beat of fusion of ice = 3.36xx 10^(5)J kg^(-1) and latent head of vaporization of water = 2.26 xx 10^(6)J kg^(-1)

What is meant by saying that the latent heat of fusion of ice is 3.34 xx 10^(5)J//kg ?

How should 1 kg of water at 50^(@)C be divided in two parts such that if one part is turned into ice at 0^(@)C . It would release sufficient amount of heat to vapourize the other part. Given that latent heat of fusion of ice is 3.36xx10^(5) J//Kg . Latent heat of vapurization of water is 22.5xx10^(5) J//kg and specific heat of water is 4200 J//kg K .

What is meant by saying that the latent heat of vaporisation of water is 22.5 xx 10^(5) J//kg ?

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1 kg of ice at 0^(@)C is mixed with 1.5 kg of water at 45^(@)C [latent heat of fusion = 80 cal//gl. Then

In a mixture of 35 g of ice and 35 g of water in equilibrium, 4 gm steam is passed. The whole mixture is in a copper calorimeter of mass 50 g . Find the equilibrium temperature of the mixture. Given that specific heat of water is 4200 J/kg K and that of copper is 420 J/Kg K and latent heat of fusion of ice is 3.36xx10^(5) J/kg and latent heat of vaporization of water is 2.25xx10^(6) J/kg.

PHYSICS GALAXY - ASHISH ARORA-HEAT AND THERMAL EXPANSION-PRACTICE EXERCISE
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