1
$\begingroup$

I'm wondering about the most efficient way of producing ice from water at room temperature.

First of all: why, in principle, do we need to put in work to do so, if water gives off energy while cooling? I could say that when it's at equilibrium with its surroundings, the entropy of the universe is maximized, so in order to change that (= lower the entropy) we need to put in work, but I don't quite understand why exactly would entropy be lower at non-equlibrium.

Anyway, we know that what refrigerators do is exactly that: they put in work in order to draw some heat from a system and dump it outside. The efficiency of a refrigerator, unlike a heat engine, has only a lower cap, so theoretically, one could think of a refrigerator with such a high efficiency, that work required to cool something would approach 0. Presumably however, creating such a fridge (for instance, with an extra cold interior) would require work on its own. Another way of freezing the water at room temperature would be to put it in a container and depressurize it, so that the water would boil and lose energy through evaporation, eventually freezing. I'm sure one could devise other methods.

But what would be the theoretical minimal work required to do so? When it comes to chemical reactions, the net work required to ignite one or net work available to acquire from one if it's spontaneous is given by the Gibbs free energy. But while it's easy to calculate for gases, I'm not sure how one would go about applying it here.

$\endgroup$
  • $\begingroup$ Your example of depressurizing the container of water occurs by removing mass from your system, which is accompanied by removing internal energy from the system. So it is not a closed system process. $\endgroup$ – Chet Miller May 14 at 15:23
3
$\begingroup$

Let's say you want to completely freeze a mass $M$ of water. The total heat removed from the water is the latent heat (ignore the superheat unless you start at a very high temperature):

$$ Q = Mh_{sf}$$

where $h_{sf}$ is the enthalpy of fusion. This is the required refrigeration load. The work required by the refrigerator is

$$ W = \frac{Q}{COP}$$

and this is minimum when the coefficient of performance $COP= \frac{T_C}{T_H-T_C}$ (i.e., $COP$ is maximum). So we have

$$ W = \frac{Mh_{sf}(T_H-T_C)}{T_C}$$

which is the minimum work required to freeze a mass $M$, using a refrigeration cycle.

$\endgroup$
  • $\begingroup$ excellent summary. $\endgroup$ – niels nielsen May 14 at 2:24
  • $\begingroup$ So the minimal work requires the refrigerator to work with Th = room temperarure and Tc just slightly below 0 Celsius? $\endgroup$ – neverneve May 14 at 9:46
1
$\begingroup$

"Minimum work required" / "maximum work available" questions can be answered using the concept of exergy, which considers both the state of the system and the state of its surroundings. For a closed system,

$$ W_\text{in} \geq X_2 - X_1 \quad\quad \text{or} \quad\quad W_\text{out} \leq X_1 - X_2 $$ where $X_2$ and $X_1$ are the system exergy at the final and initial states respectively,

\begin{align} X \equiv m (u - T_0 s + P_0 v ), \end{align} and $T_0$ and $P_0$ are the properties of the surroundings (assumed fixed).

It follows that, in this case \begin{align} W_\text{in,min} &= X_2 - X_1 \\ &=m (u_2 - u_1 + T_0(s_2 -s_1) + P_0(v_2 - v_1)), \end{align} where

  • subscript $0$ denotes properties of the surroundings
  • subscript $1$ denotes properties of the liquid water, and
  • subscript $2$ denotes properties of the ice.

This approach works even if states $1$ and $2$ are at different temperatures or pressures from each other or from the surroundings. It is based on the idea that any expansion/contraction will necessarily involve work transfer to/from the surroundings at $P_0$ and that heat transfer could, in the best case, be achieved by connecting to the system to the surroundings at $T_0$ with a reversible Carnot device and supplying/extracting the necessary work. Since this model process would be reversible, no other process could require less/extract more work.

Note that (in my experience) exergy is known in mechanical engineering, but not in chemical engineering or physics, so depending on what other sources you are consulting you may or may not be able to find additional information about this techniques.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.