# Reversible heat exchange between water and ice

Tags:
1. Jun 28, 2016

### Soren4

1. The problem statement, all variables and given/known data
In an adiabatic container are placed , in rapid succession , a mass of ice , $m_I= 2 kg$ , at temperature $T_I = -10 ◦C$ and a mass of water , $m_W = 1 kg$ , at the temperature $T_W = + 20 ◦C$ . Determine :
a) the total mass of water present in the container at equilibrium ;
b) the variation of thermodynamic entropy of the universe in the process .

Also determine :

c ) the total mass of water present at equilibrium in the case in which the equilibrium condition is achieved through a series of infinitesimal and reversible heat exchanges, knowing that the equilibrium temperature is $T_0 = 0 ◦C$ .

2. Relevant equations
$\Delta U_{system}=0$ for isolated system, $\Delta S_{system}=0$ for reversible adiabatic

3. The attempt at a solution
I'm ok with point a) and b), I just wanted to be sure about one thing. In this case the heat given by water to go to $0 °C$ is bigger than the heat absorbed by ice to go to the same temperature, but not enough to change all the ice in water. Is it correct to think that, once water reaches $0 °C$ (end some ice melted) then there would be no heat exchange any more?

The main doubt is about point c). I intrerpreted the question in terms of entropy, that is, the entropy of the system (which is isolated) does not change

$$\Delta S= m_I c_I ln(\frac{T_0}{T_I}) +m_W c_W ln(\frac{T_0}{T_W}) + \frac{\Delta m_{I} \lambda_I}{T_{0}}$$

Where $\Delta m_{I}$ is the mass of ice melted and $\lambda_I$ is the entalpy of melting ice. The only thing I'm not ok with is that, with this method I assumed (again) that water do not freeze, so that the situation is the same as the one of point a) and b) but the condision is not $\Delta U_{system}=0$ but $\Delta S_{system}=0$.

Is it correct to do so and is there a way to be sure that water cannot freeze in this situation?

Last edited: Jun 28, 2016
2. Jun 28, 2016

### Staff: Mentor

I don't understand what is being asked in part c, but please show us your work for parts a and b.

Chet

3. Jun 28, 2016

### Soren4

Thanks for the answer, part c) is the same of a) (find the mass of the water present in Equilibrium conditions) but in the hypotesis of reversible process.

For part a) I did the following
Since the $|m_W c_W (0°C-T_W)|> m_I c_I (0°-T_I)$ ice melts but, since $|m_W c_W (0°C-T_W)| - m_I c_I (0°-T_I) < m_I \lambda_I$, just some part of the ice $\Delta m_I$ becomes water and final temperature must be $0°C$. So I wrote $\Delta U_{system}=0$ to calculate $\Delta m_I$

$$m_W c_W (0°C-T_W) + m_I c_I (0°C-T_I)+\Delta m_I \lambda_I=0$$

For part b) I calculated entropy as

$$\Delta S_{system}= m_I c_I ln(\frac{0°C+273.15}{T_I}) +m_W c_W ln(\frac{0°C+273.15}{T_W}) + \frac{\Delta m_{I} \lambda_I}{0°C+273.15}$$

I wonder if water really do not freeze and if, for point c), is correct to impose $\Delta S_{system}=0$

4. Jun 28, 2016

### Staff: Mentor

I agree with your analyses of parts a and b. Nice job.

I still don't understand the question in part c. Is that the exact wording of the question?

5. Jul 8, 2016

### Soren4

Thanks for the reply! I'm a little in late! Yes I translated literaly.. I'm quite sure it is asking to consider the same process of point a) and b) but in the case that process is reversible (in fact the heat exchanges are infinitesimal) and calculate the mass of water present at equilibrium.

What I'm less sure about is if it is correct - under the hypotesis of reversibility of the process - to proceed as I did and impose $\Delta S_{system}=0$ to get $\Delta m_I$ (and therefore the mass of water present at equilibrium), that is

$$m_I c_I \mathrm{ln}(\frac{273.15 K}{T_I}) +m_W c_W \mathrm{ln}(\frac{273.15 K}{T_W}) + \frac{\Delta m_{I} \lambda_I}{273.15K}=0$$
Is this the correct approach?

6. Jul 8, 2016

### Staff: Mentor

I guess. It seems to be the only thing that kind of makes sense.

7. Jul 11, 2016

### Andrew Mason

For part c) one way to imagine how this reversible heat transfer might work would be to separate the water from the ice and run a small carnot engine between the water (as the hot reservoir) and the ice (as the cold reservoir) for an infinite number of cycles until the temperatures of water and ice were the same (each cycle drawing an arbitrarily small amount of heat from the water so as to create an arbitrarily small temperature change in the reservoirs each cycle). The work output would be kept separate and apart from the system (ie. separate from the reservoirs + engine).

Since the change in entopy of the system is 0, you can determine the heat flow into the ice by using an entropy approach.

You need to break it down into two parts. First determine the temperature of the water when the ice is brought up to 273K using entropy ($\Delta S = 0$).

$$\Delta S_{I} = \int_{263}^{273}m_Ic_IdT/T = m_Ic_I\ln\frac{273}{263} = - \Delta S_{w} = -\int_{293}^{T_{i0}}m_wc_wdT/T = m_wc_w\ln\frac{293}{T_{i0}}$$

Then use the same method to determine how much heat is delivered to the ice at 273K to bring the water down to 273K from $T_{i0}$.

AM

Last edited: Jul 11, 2016