Steam at 100^(@)C is passed into 20 g of water at 10^(@)C, then water acquires a temperating of 80^(@)C, the man 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, then water

1.	Steam at 100^(@)C is passed into 20 g of water at 10^(@)C, then water acquires a temperating of 80^(@)C, the man 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)]
Answer» 24 g 31.5 g 42.5 g 22.5 g Solution :(d) Heat gain by water = heat loss by steam `20 xx 1 xx (80 –100) = m xx 540 + m xx 1 × (100 – 80)` 1400=560 m `m= (1400)/(560)` =2.5 g Total MASS of water = 20 + 2.5 = 22.5 g