# Is it possible to harness energy from the exothermic reaction of changes of state of matter?

Since a substance emits heat as it transitions from a high energy state (gas) to a lower energy state (solid), is it possible to devise a method to capture this heat and convert it to usable energy?

I'm not sure how economical this would be, but consider the exothermic reaction of water freezing. The heat of fusion for water is $3.33 *10^5 J/kg$

So if a 1 cubic meter of water was frozen, there would be: $3.33 * 10^5 J/kg * 1000kg = 3.33 * 10^8 J$ of heat energy produced.

The heat value for the vaporization of water is $2.36 * 10^6 J/kg$ which would be even more efficient, though it is more difficult to capture 1000kg of water vapor.

I imagine this might be akin to how an air conditioner works, but I cannot imagine how this may be done or if it is already being attempted. So I pose the question here, would it be possible to harness the heat energy from the air and water concentrate it into a usable form of energy--a clean form of energy at that. Why, or why not?

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We already do harness it. –  Parth Vader Apr 10 '14 at 7:04

As you guessed in the end of your question: We already use the energy released or absorbed in phase transitions to transport energy. For example, when we put ice on a drink, and also in refrigerators and air conditioners we remove thermal energy from a region that is already cold and release it where it is already hot. The problem is that these processes do not generate energy, they only store energy that must be obtained from some source, such as burning fuel or running water downhill in a hydroelectric plant. In a geothermal vent we might find large amounts of water that is being transformed into vapor and this can be used to generate energy, as long as there is a natural source of heat to replenish the part that we used.

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Nothing, generates energy per se. But consider for a moment if the same principals of an air conditioning system were used to transfer heat from the ocean to a boiler which powered a steam engine to generate electricity. The evaporator unit in the ocean to store the thermal energy, the condenser unit in the boiler to gather the thermal energy and then concentrate it to power a boiler so the thermal energy is converted into usable energy. –  Klik Dec 9 '13 at 19:33
Yes, but a condenser takes power to run; where does that come from? –  Asher May 8 at 12:35

This has already been tried. There is an implication in the problem statement that you want to convert heat to work. This can easily be done, but the thermodynamic description of a heat engine says that the efficiency of the conversion from heat to work depends on the temperature of the heat source and the temperature at which the waste heat is rejected. The bigger the temperature difference between the heat source and heat sink, the larger the conversion efficiency from heat to work, according to the equation

$$eff = 1 - \frac{T_c}{T_h}$$

where temperatures are absolute temperatures (normally Kelvins). This means that for the melting ice process described, the efficiency will be zero if you take your heat source as melting ice and declare your heat sink to be melted water at zero deg C. There will be a relatively low (but non-zero) efficiency if your heat is supplied by condensing steam at 1 atm of pressure and your heat sink is melting ice.

Normally, you don't have control of the temperature of the heat sink (it is usually the ambient temperature), but you do have control over the temperature of the heat source. Thus, to maximize efficiency, you want to maximize the temperature of the heat source, within the ability of your equipment to tolerate the operating temperature.

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Latent heat cannot be used to extract energy, it is the change in the Gibbs free energy that you can convert to useful work. This is easy to understand intuitively, if you have a liquid in equilibrium with its vapor, then the Gibbs energy change due to an amount of the substance changing phase is zero, therefore you cannot extract any useful energy from this process no matter how much latent heat is released in this process.

This also means that you can extract work from a process that requires a lot of heat as long as the Gibbs energy drops and hence it's a spontaneous process. E.g., despite the fact that it costs a lot of energy to evaporate water, you can still extract a lot of energy from this process, simply because its a spontaneous process. So, let's calculate how much energy you can extract from 1 liter of water at room temperature at some relative humidity of r.

The 1 liter of water would be in thermal equilibrium had the humidity been 100%, therefore the Gibbs energy of the water is the same as that of water vapor at the partial pressure corresponding to 100% humidity. The drop in the Gibbs energy when all the water evaporates at the humidity of r is thus the same as the difference between the Gibbs energy of water vapor at a partial pressure corresponding to 100% humidity and at r. Now water vapor is to a good approximation an ideal gas, this allows us to calculate the Gibbs energy difference quite accurately.

From the fundamental thermodynamic relation:

$$dE = TdS - P dV$$

It follows that the differential of $G\equiv E - T S + P V$ is given by:

$$dG = -SdT + V dP$$

If we have an ideal gas at temperature T and pressure $P_1$ and we change the pressure to $P_2$ while keeping the temperature the same, then it follows the above equation that:

$$G(T,P_2)-G(T,P_1) =NKT\int_{P_1}^{P_2}\frac{dP}{P} = NkT\log\left(\frac{P_2}{P_1}\right)$$

The ratio of the pressures $\frac{P_2}{P_1}$ is the relative humidity $r$ when we take $P_1$ to be the partial pressure of water vapor at 100% humidity and $P_2$ the partial pressure at a fraction r of this. So, the maximum amount of work, which is given by the drop in the Gibbs energy, is given by:

$$W = -NkT\log(r)$$

For 1 liter of water at room temperature and $r = 0.6$ this is about 70.3 kJ which is about 1.65 times the amount of energy in my fully charged camera battery.

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