[" fleam at "100^(@)C" is passed into "20gm" g water "],[10^(@)C" .Find temp.of mixture is "80^(@)C" ."],[" of water present wall be "]

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) ]

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) ]

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

Steam at 100^(@)C is passed into 20g of water at 10^(C when water acquires temp of 80^(@)C the mass of water present will be (take specifies heat of water 1 cal g^(1) c^(-1) & latent heat of steam =540 cal g^(-1) c^(-1)

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 )

If 2g ice at 0^(0)C is dropped into 10g water at 80^(0)C then temperature of mixture is

Steam at 100^(@)C is passed into 22 grams of water at 20^(@)C . When resultant temperature is 90^(@)C , then weight of the water present is