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

Latent heat of fusion of lead is 5"cal"/(gm) and its melting point is 328^@C .How much heat is required to melt 5 gm of lead initially at 18^@C ? (Specific heat of lead = 0.031"cal"//(gm)//^@C) .

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 ).

20 g of steam at 100^@ C is passes into 100 g of ice at 0^@C . Find the resultant temperature if latent heat if steam is 540 cal//g ,latent heat of ice is 80 cal//g and specific heat of water is 1 cal//g^@C .

In a container of negligible heat capacity, 200 gm ice at 0^(@)C and 100gm steam at 100^(@)C are added to 200 gm of water that has temperature 55^(@)C . Assume no heat is lost to the surrpundings and the pressure in the container is constant 1.0 atm . (Latent heat of fusion of ice =80 cal//gm , Latent heat of vaporization of water = 540 cal//gm , Specific heat capacity of ice = 0.5 cal//gm-K , Specific heat capacity of water =1 cal//gm-K) Amount of the steam left in the system, is equal to