Solved thermodynamics problem - Can you interpret the solution, please?

[playmannn]

Homework Statement

"In a closed container there is water in equilibrium with saturated vapor at temperature 100°C (212°F). The ratio of the masses vapor/water is β=0.1. The heat capacity of water is C(o) = 4.2 J/g.K. Find the specific heat capacity of the system, considering that the specific heat of transition is q=2.26 J/g. Identify the specific heat capacity Cv of a thin fog (saturated vapor with water drops, the full mass of which is much lower of the mass of the vapor)."

Homework Equations

The solution with the key steps is provided in the attachment!

The Attempt at a Solution

Can you, please, tell me how are the stated equations derived, starting with why is the one for mc like so, as well as the following ones. I'll be so thankful if you do!

 

[mark726]

Thank you for your question. I am happy to explain the steps and equations used in the solution provided. First, let's define some variables:

- m: mass of water
- m_v: mass of saturated vapor
- m_f: mass of fog (saturated vapor with water droplets)
- C: specific heat capacity of the system (water and vapor)
- C_v: specific heat capacity of fog
- T: temperature of the system
- q: specific heat of transition (latent heat)

The key equation used in the solution is the heat capacity formula:

Q = mCΔT

Where Q is the heat absorbed or released, m is the mass of the substance, C is the specific heat capacity, and ΔT is the change in temperature.

In this problem, we are given the temperature of the system as 100°C, which is also the boiling point of water at standard atmospheric pressure. This means that the water and vapor are in equilibrium, and the heat absorbed by the system is used to convert water into vapor. This is known as the latent heat of vaporization, and it is represented by the variable q.

Now, we can write the heat capacity equations for the water and vapor separately:

Q_w = mC_wΔT_w

Q_v = m_vC_vΔT_v + m_vq

Where ΔT_w and ΔT_v are the change in temperature for water and vapor respectively. Since the system is in equilibrium, we can assume that the temperature of both water and vapor are equal to the boiling point of water, 100°C. Therefore, ΔT_w = ΔT_v = 100°C.

We are also given the ratio of masses of vapor to water, β=0.1. This means that for every 1 gram of water, there is 0.1 grams of vapor. We can use this ratio to find the mass of vapor, m_v:

m_v = βm = 0.1m

Substituting this into the heat capacity equation for vapor, we get:

Q_v = (0.1m)C_v(100°C) + (0.1m)q

Now, we can equate the heat absorbed by the water to the heat released by the vapor, since the system is in equilibrium:

Q_w = Q_v

mC_wΔT_w = (0.1m)C