Question

A sample of steam with a mass of 0.507 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.45 g of water at 4.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C) Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

In the given question, the changes involved are:

A) The first involves condensation of the steam (at 100°C) to liquid water at 100°C (boiling point of liquid water is 100°C).

B) next is the cooling of liquid water from 100°C to the final equilibrium temperature.

C) The other change is the heating of the liquid water in the container to the equilibrium temperature.

To reach the equilibrium temperature, the heat lost by steam and the hot water formed by it is gained by the cool water in the container. Obviously, this considers that none of the heat lost/gained is transferred to the surroundings.

Thus, for equilibrium, q_A + q_B = q_c ( only the magnitudes of heat must be taken, since it’s already mentioned as ‘heat lost’ / ‘heat gained’. The sign of q is important only when not specifying whether heat is gained or lost).

Change A :

To find q, multiply the no. of moles by the enthalpy of condensation

Also, condensation = opposite of vapourisation

So, enthalpy of condensation = (-) enthalpy of vaporisation = - 40.7 kJ/mol

Mass of steam = 0.507 g. Molar mass of water = 18g/mol

So, number of moles = Mass/Molar mass = (0.507 g)/(18 g/mol) = 0.0282 moles

So, q_A = 0.0282 moles x (-40.7 kJ/mol) = - 1.1464 kJ = -1146.4 J

Here, q_A represents heat change. So, the negative sign denotes heat lost to surroundings.

Change B:

The formula which we need to use is:

q = mC_pΔT

where q = heat transferred (in J), m = mass, Cp = specific (in J/(g°C)), and ΔT = Final – Initial temperature (in °C)

Putting the values:

m = 0.507 g

Cp(we take for liquid water) = 4.18 J/(g. °C)

Let final equilibrium temperature = x.

ΔT = x°C – 100°C = (x-100)°C . ( since x will be less than 100, so, (x-100) = negative, making heat change negative)

So, q_B = 0.507 g x 4.18 J/(g. °C)x (x-100)°C .

Taking (-) sign out: q_B = -[0.507 g x 4.18 J/(g. °C)x (100 -x )°C]. This is done so that when the magnitude of qB is put in the equation later, it will represent the amount of heat lost to the surroundings.

Change C:

q = mC_pΔT

For cold water:

m = 4.45 g. Cp = 4.18 J/(g. °C) , ΔT = x°C – 4.0°C = (x-4)°C . ( since x will be more than 4, so, (x-4) = positive, making heat change positive. )

So, q_C = 4.45 g x 4.18 J/(g. °C)x (x-4)°C .

For equilibrium,

q_A + q_B = q_c

1146.4 J + [0.507 g x 4.18 J/(g. °C)x (100-x)°C] = [4.45 g x 4.18 J/(g. °C)x (x-4)°C]

Simplifying :

1146.4 J + 2.119(100-x) J = 18.601 ( x-4) J

Solving for x, x = 69.14

Thus, final equilibrium temperature of mixture = 69.14°C