Steam is passes into 22 g of water at `20^@C`. The mass of water that will be present when the water acquires a temperature of `90^@C` (Latent heat of steam is `540 cal//g`) is

Steam is passed into 54 gm of water at 30^(@)C till the temperature of mixture becomes 90^(@)C . If the latent heat of steam is 536 cal//gm , the mass of the mixture will be

Steam is passed into 54 gm of water at 30^(@)C till the temperature of mixture becomes 90^(@)C . If the latent heat of steam is 536 cal//gm , the mass of the mixture will be

Steam at 100^@C is passed into 20 g of water at 10^@C When water acquires a temperature of 80^@C , the mass of water present will be Take specific heat of water= 1 cal g^-1^@ C^-1 and latent heat of steam = 540 cal g^-1 )

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C when water acquire a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water = 1 cal g^(-1).^(@) C^(-1) and latent heat of steam = 540 cal g^(-1) ]

Steam at 100^(@)C is passed into 20 g of water at 10^(@)C when water acquire a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water = 1 cal g^(-1).^(@) C^(-1) and latent heat of steam = 540 cal g^(-1) ]

Stream at 100^(@)C is passed into 20 g of water at 10^(@)C . When water acquires a temperature of 80^(@)C , the mass of water present will be [Take specific heat of water =1 cal g^(-1) .^(@)C^(-1) and latent heat of steam =540 cal g^(-1) ]