Steam at 100 degrees Celsius is bubbled into 250 cm3 of water at room temperature in a calorimeter cup. How much steam will have been added when the water in the cup is at 60 degrees Celsius? (Ignore the effect of the cup.) So I did: -(Heat lost by steam) = Heat gained by water msteamLvap + msteamc(40[degrees]C) = mwaterc(40[degrees]C), where c = specific heat of water solving for msteam, i get 1.72 * 10^-2 kg. But the answer is 1.79 * 10^-2 kg. What's wrong with my calculation?