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

The amount of energy transferred can be calculated using the equation:

Q = mc ΔT

Where:

Q = the heat energy transferred (joule, J)

m = the mass of the liquid being heated (grams, g)

c = the specific heat capacity of the liquid (joule per gram degree Celsius, J/g°C)

ΔT = the change in temperature of the liquid (degree Celsius, °C)

L(fusion)=333.5J/kg
L(vaporisation)=2257J/g

Also Q=mL ; where Q is the energy required to change state and L is either latent heat of fusion (melting) or vaporisation (condensation)

The total energy required to raise the temperature of the system 25°C. This will be the sum of the energy required to heat 2.20 kg of water by 25°C, the energy required to melt 0.450 kg of ice and the energy required to heat 0.450 kg of water by 25°C.

Energy required to heat 2.20 kg of water by 25°C: Q=2.20*4200*25=231000 J
Energy required to melt 0.45 kg of ice: Q=0.45*(333.5*10^3)=150075 J
Energy required to heat 0.45 kg of water by 25°C: Q=0.45*4200*25=47250 J
Therefore total energy required is Q=428325 J

Then we need to work out how much steam needs to condense to provide this energy.

m=Q/L = 428325 /2257= 189.77 g

Thus 189.77 grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 25°C.