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`.

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 .

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' ""^(@) 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 heat of water = 1 cal g^(-1) ""^(@) C^(-1) latent heat of fussion of ice = 80 cal g^(-1) , Latent heat of steam = 540 cal g^(-1) ]

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) ].

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

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

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 )

How much of ice will be converted into water at 0^@C if 10g of steam condenses into water at on a block of ice ? (latent heat of ice = 80cal/g, latent heat of steam = 540cal/g).