Steam at 100   o C is passed into 20   g of water at 10   o C . When water acquires a temperature of 80 o C , the mass of water present will be: [Take specific heat of water = 1   c a l   g - 1   o C - 1 and latent heat of steam = 540   c a l   g - 1 ]

Step-by-step solution

Given: Specific heat of water S w = 1   cal   g - 1 ° C - 1 , Latent heat of steam L s = 540   cal   g - 1 Let the mass of steam = m   g So heat lost in change of state from steam to water equal to sum of latent heat and heat use in raise of the temperature of water. Q 1   = m L s   + m s w ∆ T Q 1 = m × 540   + m × 1 × ( 100 - 80 ) Q 1   = 540 m   + 20 m   = 560 m Heat gained by water to change its temperature Q 2 = m w S w ∆ T