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How should 1 kg of water at 5^@C be divi...

How should 1 kg of water at `5^@C` be divided into two parts so that if one part turned into ice at `0^@C`, it would release enough heat to vapourize the other part? Latent heat of steam`=540 cal//g` and latent heat of ice `=80 cal//g`.

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The correct Answer is:
`864g+136g`
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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 .

M' kg of water 't' 0^@ C is divided into two parts so that one part of mass 'm' kg when converted into ice at 0^@ C would release enough heat to vapourise the other part, then (m)/(M) is equal to [Specific heta of water = 1 cal g^(-1) .^@ C^(-1) , Latent heat of fusion of ice = 80 cal g^(-1) , Latent heat of steam = 540 cal g^(-1) ].

How should 2kg of water at 60^(@)C be divided so that when one part of it is turned into ice at 0^(@)C it would give out sufficient heat to vaporize the other part? (Specific latent heat of fusionof ice =336xx10^(3)J kg^(-1) and specific latent heat of vaporization of steam =2250xx10^(3)Jkg^(-1) )

1 g of a steam at 100^(@)C melt how much ice at 0^(@)C ? (Length heat of ice = 80 cal//gm and latent heat of steam = 540 cal//gm )

Find the heat required to convert 1 g of ice at 0°0C to vapour at 100°C. (Latent heat of ice = 80 cal/g and latent heat of vapour = 536 cal/g)

100 gm of water at 20^@C is converted into ice at 0^@C by a refrigerator in 2 hours. What will be the quantity of heat removed per minute? Specific heat of water = 1 cal g^(-1) C^(-1) and latent heat of ice = 80 cal g^(-1)

What amount of ice will remains when 52 g ice is added to 100 g of water at 40^(@)C ? Specific heat of water is 1 cal/g and latent heat of fusion of ice is 80 cal/g.

1 g of steam at 100^@C and an equal mass of ice at 0^@C are mixed. The temperature of the mixture in steady state will be (latent heat of steam =540 cal//g , latent heat of ice =80 cal//g ,specific heat of water =1 cal//g^@C )

What will be the final temperature when 150 g of ice at 0^@C is mixed with 300 g of water at 50^@C . Specific heat of water =1 cal//g//^@C . Latent heat of fusion of ice =80 cal//g .

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