Question

A vessel whose walls are thermally insulated contains 2.20 kg of water and 0.450 kg of ice, all at a temperature of 0.0 ∘C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water.

Part A How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 25.0 ∘C ? You can ignore the heat transferred to the container.

Answer 1

Enthalpy of fusion of steam at 100C and 1 at. = 226e3 J/kg = 2.26e6 J/kg
Enthalpy of fusion of ice at 0C = 334 kJ/kg = 3.34e5 J/kg
Specific heat of water betw 0 and 100C at 1 at. = 4.19 J/g = 4.19e3 J/kg

Let m = mass of steam required, in kg. Then

Q yielded by steam = m*2.26e6 L

If we had just added m kg of water at 100C we would write
Heat lost by m kg water at 100C = heat needed to melt 0.45kg of ice + heat needed to raise the temp. of (2.2+0.45)kg of water at 0C to 25 C:

m*(100-25)*4.19e3 = 0.45*3.34e5 + (2.2+0.45)*25*4.19e3

However, to the LHS of the equation we need to add the Q due to the enthalpy of fusion of steam = m*2.26e6 J:

2.26e6m + 293300*m =150300 + 277588
2.55e6m = 427888
m = 0.168 kg steam