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

Latent heat of 1 gm of steam is 536 cal/gm , then its value in joule/kg is

One gram of ice at 0^(@)C is added to 5 gram of water at 10^(@)C . If the latent heat of ice be 80 cal/g, then the final temperature of the mixture is -

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

How much steam at 100^(0)C is to be passed into water of mass 100g at 20^(0)C to raise its temperature by 5^(0)C ? (latent heat of steam is 540cal/gm )

50 g of steam at 100^@ C is passed into 250 g of at 0^@ C . Find the resultant temperature (if latent heat of steam is 540 cal//g , latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g -.^@ C ).

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 )

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 )