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I'm trying to invent a distillation apparatus that runs solely on electricity. Suddenly, I realized that cooling things is really hard, while heating them up is so easy.

Actually, it seems that there are just three ways to cool something down:

  1. Peltier modules (incredibly ineffective)
  2. Compressing and expanding gasses (hard to make at home, the device is too big)
  3. Some rare endothermic reactions, such as dissolving $KNO_3$ in water

My question, however, is not how to solve my issue. I want to know why there are so limited cooling options and why they are so expensive and tricky.

For heating, the options are much easier:

  1. Current flow (just pick a wire and a battery)
  2. Rubbing things
  3. Burning/dissolving acids in water and other chemistry (if you're lucky, you get so much heat you won't need any more in your life)
  4. Absorbing el-mag waves

I, for one, blame the second law of thermodynamics.

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  • $\begingroup$ yes i was also wondering about cooling. There is also heat transfer like in an AC(after expansion/phase transfer), magnetic cooling and stimulated emission. but what about more ways? $\endgroup$ – ChemEng May 24 at 23:49
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It is because of the second law of thermodynamics. There are many irreversible processes that can be used to heat something. It is the natural flow of things because entropy will increase in isolated systems, and much of the internal energy of objects can be dissipated as heat (and this heat used to heat something that is colder). However, to cool something you need to perform work in order to decrease the entropy of the subsystem. There is a maximum efficiency for work to be used to cool something by removing heat from one source at lower temperature and move it to a source at a larger temperature. So both processes, heating and cooling are not symmetrical in our universe with a thermodynamic arrow of time.

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  • $\begingroup$ +1 Correct but "as far as I know you need to use a reversible process": no, the cooling process does not need to be reversible; it is just less efficient (requires more work) to compensate for both the entropy decrease of the cooled subsystem together with the irreversible entropy production by the irreversible cooling process. $\endgroup$ – Selene Routley Nov 5 '14 at 13:38
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    $\begingroup$ I feel like this answer is sort of begging the question. $\endgroup$ – DanielSank Nov 5 '14 at 16:26
  • $\begingroup$ @WetSavannaAnimalakaRodVance You are right, I corrected it, thanks! $\endgroup$ – Wolphram jonny Nov 5 '14 at 16:55
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You could blame the laws of thermodynamics and say that cooling is much harder in our universe because of them. However, since we're in a dark energy-dominated universe that's expanding and cooling, it seems as though cooling is generally easier for the universe on the largest scales.

Even on smaller scales, cooling is usually easier (I've of course determined ease by observing which one happens more often). The cores of planets will cool and harden over time, stars use up their fuel and cool as they die. Cooling by far is the easier process. Even when the universe reaches a heat death (assuming it ever does), the expansion will continue to cool it to lower temperatures. So cooling is definitely easier for the universe.

So why isn't cooling easier for us? Well the answer could be due to thermodynamics. It could be that on short time and distance scales, it is easier to heat than to cool, but let me present a more anthropological reason.

Throughout human history, we have always strove to perform tasks. We want to build buildings, grow crops, light our homes. All these tasks require us to expend or use energy. As such, we have invented brilliant systems and processes of generating easy-to-use energy and channeling it where we need it to be. We have become very adept at taking energy from a few common sources and dumping it into wherever or whatever we want. And, as we all know, putting energy into an object is much the same as increasing its temperature. So, for us, increasing the temperature of something is no problem. It's generally what we do. However, to decrease the temperature of an object, you need to remove energy from it. Now with the exception of those few specific sources, we aren't very adept at taking energy out of something. That has never really been as necessary in history because usually when you remove energy, it makes it harder to do any tasks. What would the point be of making it harder to perform tasks? As such, we rely mostly on natural processes to remove the energy from systems. But unlike the processes we invented, natural processes usually try to bring temperatures to thermal equilibrium. Sure, we invented refrigeration and we found a few endothermic reactions to exploit. But when it comes down to it, we're much more interested in putting energy into things (computers, lights, heating systems, anything requiring electricity) than taking it out.

It could simply be the case that while the universe finds cooling to be easier, we have put a lot more effort into figuring out how to heat things and so that is easier for us.

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  • $\begingroup$ I would just say that I think that the universe isn't cooling because it expands - the sum of energy doesn't change. Remember the thought experiment where there's vacuum in one part of the tube and gas in other - when you remove the separator, the temperature is constant. $\endgroup$ – Tomáš Zato - Reinstate Monica Nov 5 '14 at 17:23
  • $\begingroup$ @TomášZato The energy density of dark energy is constant in the LCDM model. Thus, as the universe expands, the energy actually increases. But the temperature of the universe is not based on the sum of all energy. If you are talking about radiation, then the energy density falls off like $1/a^4$, which means the sum of radiative energies decreases over time in an expanding universe, which is why the blackbody temperature decreases and the universe cools $\endgroup$ – Jim Nov 5 '14 at 17:28
  • $\begingroup$ Sounds legit. What does the thermal energy turn into then? Mass? Velocity? $\endgroup$ – Tomáš Zato - Reinstate Monica Nov 5 '14 at 18:27
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    $\begingroup$ @TomášZato nothing. It's gone. Energy isn't conserved globally in this case $\endgroup$ – Jim Nov 5 '14 at 18:32
  • $\begingroup$ Interesting to read this anthropological point of view, thanks. $\endgroup$ – zundi Aug 5 '18 at 21:17
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Let me offer a different persepective on this. Cooling is not universally harder than heating. To demonstrate this, consider the following:

Suppose you have two 1 kg copper blocks, one at 200 K, and one at 400 K. Put them into direct contact, and seal them in a perfectly-insulated (no heat transfer from the walls) empty box. The 400 K block will then cool to 300 K just as readily as the 200 K block warms to 300 K. [N.B.: The 300 K final temperature is based on the simplifying assumption that the heat capacity of copper is temperature-independent between 200 K and 400 K.]

But, you might protest, aren't there many instances in which cooling is more difficult—e.g., cooling vs. heating a home? The answer is yes. But then what distinguishes the example I've offered from those that concern you? The difference is that you are specifically referring to cases in which potential energy (PE) (typically electrical energy) is used for heating or cooling.

Given this, I would suggest your question be refined as follows: Why is it harder to use PE to cool than to heat? Now I can give you an answer:

PE can be converted to thermal energy with no losses, thus achieving heating directly with 100% efficiency. E.g., I can convert electrical energy entirely to thermal energy using a resistor. [This ignores losses in getting the PE into your system, e.g., resistive losses in electrical transmission.]

However, PE cannot be directly converted into the removal of thermal energy. Instead, to use PE to cool requires running some form of heat engine. And, by the 2nd law of thermodynamics, even a perfect heat engine cannot be 100% efficient. Furthermore, all real-world heat engines operate dissipatively, and thus will be even less efficient than a perfect heat engine.

So, in summary, if you are talking about the conversion of PE to heating vs. cooling, the difference is that the former can be done directly, with 100% efficiency, while the latter always requires going through some form of heat engine, with significant attendant losses (as well as [with some exceptions, such as a Coolgardie safe] significantly greater engineering complexity).

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The rate of heat transfer depends on a temperature difference between objects, and heat transfer is always from high temperature to low temperature. When heating something with a flame, the temperature difference between the flame and the low temperature object can easily be 1000 deg C, so the rate of heat transfer is high. When cooling something, the temperature difference between the high temperature object (e.g., boiling water) and the low temperature object (e.g., ice) is often low (in this case, 100 deg C). Accordingly, it is appropriate to specify a fixed temperature difference between objects before deciding that it is easy to heat things up but difficult to cool them down, as such a statement without qualifiers tends to "compare apples to oranges".

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It is because heating and cooling happen because of different physics. Consider the following equation:

$\partial T/\partial t =c \nabla^2 T + H(x,t, T)$

Where $H(x,t)$ is the external heating. This is the source term. Cooling would be a diffusive process but heating could be nonlinear. In many cases $H(x,t) >> c \nabla^2 T$ and therefore heating is faster.

This equation also tells us what you are implying may not be true in all cases. If $H(x,t) << c \nabla^2 T$, you will have faster cooling. So it depends on the parameter regime.

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  • $\begingroup$ Why can't we have external cooling, i.e. $H<0$? Not like heating by diffusion doesn't work either ... so I don't understand the difference you're creating here. $\endgroup$ – Sanya Dec 7 '16 at 13:08
  • $\begingroup$ $H$ will be negative for a freely expanding gas because of equation of state constraints. $\endgroup$ – wander95 Dec 7 '16 at 14:04
  • $\begingroup$ You're not answering the question, just recasted it into an equation and then claim strange things. No real explanation. $\endgroup$ – thermomagnetic condensed boson May 24 at 20:38

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