Suppose I have a basin of water containing $1000$L that's at a temperature of $310$ K. Now I'd like to put this thermal energy to good use (say set a cup of tea or a hot bath), would I be able to concentrate the thermal energy of this water in to a smaller basin so that the temperature of that basin gets to be $360$ K?

Now, I know that heat pumps do this with air, by relying on evaporative cooling. But in this case I'm specifically wondering if such a system would be possible when medium one takes energy away from is water?


2 Answers 2


If all you have is a basin of water at 310K, you cannot concentrate the temperature. The water is already at thermal equilibrium, so there is no way to extract energy from it in any way. That's the nature of entropy.

Now if you have that basin of water at 310K and a heat sink at a lower temperature, such as a bunch of room temperature air at 300K, or the night sky at around 230K, you can start doing something. Now the system is not at thermal equilibrium, so there is the opportunity to extract work.

In such a case, you can indeed divide the water up into two buckets, one with a cup of water, and one with the remaining 999.9 or so liters, all at 310K. You can then create a heat engine (choose your favorite) which generates usable energy by transferring heat from the large body of water into the cooler heat sink. You can then use this energy to drive a resistive heater in the small cup of water to raise it up to a temperature worthy of brewing tea.

Your cup of tea needs to be raised in temperature by about 67K to reach boiling. If your teacup contains roughly 200mL, that means you have 999.8L of water to use as a heat source. If you transfer that heat into a 300K heat sink, that's a 10K drop. You can do some quick math to show that an engine that is just 0.1% efficient is more than enough to do the job.

  • $\begingroup$ Are you sure about your first statement? I'm under the impression that it's possible even if all I have is a basin of water at 310K, but I'd need an external energy source. $\endgroup$
    – Allure
    Dec 19, 2017 at 0:24
  • 1
    $\begingroup$ @user3727079 Yes, I am sure of my first statement. If you have an external energy source, then a basin of water at 310K is not "all you have." You also have an energy source. $\endgroup$
    – Cort Ammon
    Dec 19, 2017 at 0:27

heat pumps do not rely on evaporative effects. in a heat pump, mechanical work is used to extract heat from a cold source and move it to a warm sink (think: refrigerator running backwards). Heat pumps can and are used to extract heat from a cold body of water and move it into a warmer space; in all such cases the amount of heat dumped into the warm sink is equal to the sum of the mechanical work performed by the pump plus the heat that it extracted from the cold source. so the answer to your question is Yes, you can use a heat pump to pull enough heat from the 1000L reservoir to raise the temperature of the smaller basin to the desired temperature.


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