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Let's say I have a volume $V$ filled with water of temperature $T$. Now I remove a mass of water, $\Delta m$, and I want to know how this affects the temperature in V at first order neglecting all other heat transfer in and out of the volume and other fancy stuff. I would take a simple form of the equation of energy balance so that temperature change in $V$ reflects the energy $Q$ that is taken out:

$$ V\rho c \Delta T = Q $$

where $\rho$ is the density and $c$ is the specific heat capacity.

Now I have 3 questions.

1) Does the removal of the mass have any effect on temperature at all?

If yes, 2) Can I assume that the energy of the removed mass is

$$ Q = c \Delta m T $$

or is this wrong? If it's possible, then 3) do I use temperature in °Celsius or in Kelvin?

$c$ is sometimes reported with °C and sometimes with K, because, as I understand it, it usually relates to a temperature difference rather than absolute temperature (meaning it doesn't make a difference). Used with an absolute temperature (as in this case, unless I'm wrong) it does make a difference though, so I'm not sure how to handle this here.

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  • $\begingroup$ Why would removing mass change the temperature? If it's water, it'll change the volume, as liquid are usually considered incompressible, but not the temperature. $\endgroup$
    – Martino
    Commented May 24, 2017 at 15:33
  • $\begingroup$ Why would you assume that the temperature is changed? You do not "remove" internal energy from the rest of the system, just from the part you take away. $\endgroup$
    – nasu
    Commented May 24, 2017 at 15:34
  • $\begingroup$ I added this as an additional question, thanks. Actually it is not my assumption but that in research article that I am reading. The reasoning is that the removed mass has a certain energy that is taken into account in the energy balance. $\endgroup$
    – ye-ti-800
    Commented May 24, 2017 at 15:41
  • $\begingroup$ This is really just a small point but it's worth noting that it depends what you mean by "system" in the question. If your system is the total mass, then even if there is no net heat transfer and "other fancy stuff" on the system there still may be interactions between the removed mass and the bulk which can cause the temperature of the bulk/removed mass to change. $\endgroup$
    – alex
    Commented May 24, 2017 at 18:41
  • $\begingroup$ It is quite plausible if you're physically separating the mass since most systems have some finite pressure that the bulk mass will do some work (through the pressure force) on the physically displaced mass and lose energy and thereby the temperature may change. If by "system" you meant only the bulk mass which was't removed, then of course your assumptions would imply that you weren't considering this. $\endgroup$
    – alex
    Commented May 24, 2017 at 18:44

1 Answer 1

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Let's consider what will happen when we remove that mass.

For starters, we assume this has homogeneous temperature.

Now, the first thing that seems to be confusing you may be how they define heat energy. It's relative. You are not measuring the absolute heat of the water, you are measuring the heat added (or removed) when going from one temperature to another ($\Delta T$).

You said we are not considering the heat transfer. When you remove a mass, you aren't replacing it with cold water; you're just not considering it as part of your system anymore.

Your value of $Q$ (heat) will change, because you now have less of the warm object. The value of $\Delta T$ will remain the same; because you didn't allow for any heat transfer to lower the temperature.

Lets say you had 2 apples sitting beside each other at the same temperature. They have a heat energy associated with their temperature, mass, and heat capacity ($Q=mc\Delta T$). If you take away one apple, consider what happens to the remaining apple. It does not suddenly lose temperature just because there is no other apple there. Instead, the whole system has less thermal energy ($Q$) due to the lower mass. $\Delta T$ remains constant.

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  • $\begingroup$ Does it make a difference if you let the removed water be replaced by air (of same temperature), instead of just removing the water? $\endgroup$
    – ye-ti-800
    Commented May 24, 2017 at 16:35
  • $\begingroup$ @ye-ti-800 No, if the air is the same temperature as the water, there will be no heat transfer between the two and $Q$ will not change, therefore $\Delta T$ won't change. $\endgroup$
    – JMac
    Commented May 24, 2017 at 16:39
  • $\begingroup$ Let's say somebody uses this equation and plugs in $T=40°C$ into $Q=c\Delta m T$. Would this imply that the removed water is replaced by water which is 40° cooler? $\endgroup$
    – ye-ti-800
    Commented May 24, 2017 at 18:46
  • $\begingroup$ @ye-ti-800 $Q=c\Delta m T $ wouldn't really work in the way we traditionally use the equation. You can't really use it without a $\Delta T $ because you're comparing the energy between two temperatures. Its not a measure of absolute energy. You need a $\Delta T $ if you want to use that equation. One important part about the $\Delta T $ is that you can use K or °C because they scale the same, so the difference is equivalent with °C and K. The value of $c $ is formulated to work with a $\Delta T $. $\endgroup$
    – JMac
    Commented May 24, 2017 at 18:56
  • $\begingroup$ that's what my question was aiming at. I'll try to rephrase: since the 40C can not be an absolute temperature, when you use it in that equation (where it should be $\Delta T$) you actually calculate the energy change due to cooling of $\Delta m$ by 40C. Is that correct? I'm asking because I read someone using it to calculate the energy change due to removal of a $\Delta m$ with T=40C (absolute T) $\endgroup$
    – ye-ti-800
    Commented May 24, 2017 at 19:32

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