Question

What mass of ice at -10.0∘∘C must be added to 52.0 g of steam at 120∘∘C to end up with water at 40.0∘∘C? (Express your answer to three significant figures.)

Answer 1

Temperature of Ice = -10 degree celcius

Temperature of Steam = 120 degree celcius

Final temperature of water = 40 degree celcius

Thus

total amount of heat required by ice to reach 40 degree celcius is

(from -10 to 0) (at o degree celcius to phase change) (from 0 to 40)

Now the total energy lost by steam from 120 degree celcius to reaching 40 degree celcius of water

(from 120 to 100) (for phase change at 100) (from 100 to 40)

by putting all the values in above equation, we get

we know that in the give problem heat lost by steam is equal to the heat gained by ice to reach at equilibrium temperature at 40 degree celcius. Thus by equating both the energies, we get

Thus the mass of ice required to convert 52 gram of steam into water at 40 degrees is