Question

We have an unknown mass of liquid water at 30 ° C. It is heated by 18,000 calories and it produces the mixture: liquid water-water vapor. Water vapor is 40% of the mixture. The specific heat of liquid water is 1.00 cal / g ° C, and its latent heat of boiling is 539.7 cal / g. Determine the initial mass of liquid water.

Answer 1

Let m be the mass of liquid water initially.

Heat required to convert m mass of water at 30o to m mass of water at 100o(as water boils),

Q1 = mcw(100-30) = m*1*70 = 70m

Now, mass of water liquid water converted to steam = 40% of m = 0.4m

Heat required to convert 0.4m mass of liquid water at 100o to 0.4m mass of steam at 100o C,

Q2 = 0.4m*Lv = 0.4m*539.7 = 215.88 m

According to question, Q1 + Q2 = 18000

So, 70m + 215.88m = 18000

Or, 285.88m = 18000

Or, m = 18000/285.88 = 62.96

So, initial mass of liquid water = 62.96 gm