I have two objects, A and B. A and B are identical in every way except their temperature. Units are not relevant: assume degrees C, F, or anything linear with them.

Also not relevant are phase transitions. Assume that either nothing is freezing/melting/otherwise, or at least that the effects are negligible. The matter is mostly or entirely in the same phase throughout the event.

The temperature of A is 100 degrees.

The temperature of B is 200 degrees.

Applying a heat source raises the temperature of A by 10 degrees up to 110 degrees.

With no modifications, the heat source is applied to B. Again, all other things being equal: should B be expected to heat up by 10 degrees also, or by some other amount?

For an example, consider a room that you want to warm with a fire. 1) The temperature of the room is 10 degrees and the addition of the fire warms it by 20 degrees up to 30 degrees. OR 2) The temperature of the room is actually -50 degrees, but will that same fire warm the room by 20 degrees up to -30?

I ask this because I thought the temperature change would be different in the two different situations. Taking the above example, I would have thought the -50 degree room would be warmed by more than 20 degrees (25? 30? I don't know, just more than 20.) But someone is telling me that the -50 degree room would be warmed by the same 20 degrees, not more.

My thought process is along these lines: Two objects will attempt to equalize their heat when brought together, so 10 degree water added to 30 degree water will result in approximately 20 degree water, but add that same water to 50 degree water and it will result in 30 degree water instead (+/- 20 instead of 10). I realize this is not the same as a heat source, but I thought something similar would apply.

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    $\begingroup$ Unclear question. There may be a phase transition between the two temperatures. And apart from that, heat capacity depends on temperature. $\endgroup$ – Pieter Jul 12 '18 at 16:57
  • $\begingroup$ @Pieter Thank you. Edited concerning phase transition (ie: it's negligible and ignored). How do I clarify concerning your comment on heat capacity? $\endgroup$ – Aaron Jul 12 '18 at 17:03
  • $\begingroup$ @Pieter Or rather: was your comment on heat capacity intending that I need to clarify more for that as well, or was that merely informational for my benefit? (ie: "The number of heat units needed to raise the temperature 1 degree [heat capacity] depends on temperature, so no, you cannot expect the temperature change to be the same for the two separate cases") $\endgroup$ – Aaron Jul 12 '18 at 17:06
  • $\begingroup$ Your example of mixing water is totally different from having a heat source because the mixing of water changes the mass of the system which must come to an equilibrium. That's different from simply adding energy to a fixed mass. Take your 30 degree and 50 degree water samples and simply at 10 Cal/g. Both rise approximately 10 degrees. $\endgroup$ – Bill N Jul 12 '18 at 17:34

It depends if the heat transfer properties of the materials are temperature dependent or not.

For example, let's think about the heat capacity, which is a measure of how much energy is required to produce a certain change in temperature. If the heat capacity is constant at all temperatures, then for both cases the objects will have the same change in temperature. If, however, the heat capacity decreases with increasing temperature, then the hotter object will not get as much of a temperature increase as the colder object.

All of this also depends on how long the heat source is supplied and the temperature dependency on other things like conductance of the objects. If everything is independent of temperature, and if the objects are identical with regards to their material properties, then yes, the two objects will experience the same temperature change.

  • $\begingroup$ The only substance that I can think of would be a monatomic gas, sufficiently dilute to behave as an ideal gas. $\endgroup$ – Pieter Jul 12 '18 at 17:13
  • $\begingroup$ @Pieter That is fine :) The OP is not asking about any substance in particular, or whether certain types of substances exist. $\endgroup$ – Aaron Stevens Jul 12 '18 at 17:14
  • $\begingroup$ I am looking for "the normal case" that one might experience with everyday situations such as the air in their house. I understand that "the normal case" might be a naïve thing to say, as perhaps there is no such thing... I do not know. Is it common for masses that an average individual interacts with on a normal basis to have a heat capacity that changes with temperature? Uncommon? A good mixture of "constant at all temperatures" and not constant? $\endgroup$ – Aaron Jul 12 '18 at 17:16
  • $\begingroup$ To clarify my comment: Though what you say may be perfectly true, for me it is like telling someone who has zero understanding of electronics that some substances have practically zero resistance in certain situations, and having them walk away thinking that superconductors might be normal everyday things. Hence my previous comment. That is where I'm at with what you have said about heat capacity depending on temperature. $\endgroup$ – Aaron Jul 12 '18 at 17:18
  • $\begingroup$ @Aaron oh ok so you are looking for more specific examples. If you are interested in air I would look into the thermal properties of air. It all depends on the substance in question, and I am not an expert on such things. $\endgroup$ – Aaron Stevens Jul 12 '18 at 17:29

It depends on how you choose to define "same heat source".

If you are saying that the net heat transfer is the same in either scenario, then I agree with the other answers. It would all depend on the material properties and if they vary with temperature.

Your wording and example implies something a bit different to me though. You seem to be suggesting that we would be using the same heat source, at the same temperature, and then comparing the final temperatures of the heated objects. In that case; the starting temperature is very important.

This is because the rate of heat transfer is relative to temperature difference. A 400 degree campfire will raise the temperature of a 100 degree stick more than a 200 degree stick; all other factors the same. That's because the 100 degree stick will get a higher net energy input from the flame; due to the greater thermal gradient driving the heat transfer.

  • $\begingroup$ Your last paragraph was (I think) what I was thinking, as per my meta-information toward the end of the question. What prompted this question was a discussion about camping in a thick tent or igloo in Antarctica in approximately -100F weather. I suggested that your heat source (such as a camp stove) would provide a larger temperature change than what you are used to at more normal temperatures (ie: "If you're used to it warming your shelter by 10 degrees here, expect more than 10 degrees warming there, bringing you up to a "balmy" -80F instead"). $\endgroup$ – Aaron Jul 12 '18 at 18:00
  • $\begingroup$ Does my previous comment change anything? Is there some way in which this makes a difference for which I should edit the question (or ask a new question since this one already has three answers)? $\endgroup$ – Aaron Jul 12 '18 at 18:00
  • The temperature of A is 100 degrees.

  • The temperature of B is 200 degrees.

  • Applying a heat source raises the temperature of A by 10 degrees up to 110 degrees.

With no modifications, the heat source is applied to B. Again, all other things being equal: should B be expected to heat up by 10 degrees also, or by some other amount?

  1. Object A starts at 100 degrees.

    You apply a temperature of 110 degrees to A.

    Eventually object A reaches a temperature of 110 degrees because you said: "Applying a heat source raises the temperature of A by 10 degrees up to 110 degrees." - IE: your heat source does not cool or run out of fuel until object A reaches 110 degrees, at which point the heat source is removed from the equation and object A continues to heat or commences cooling to match the surrounding temperature.

  2. Object B starts at 200 degrees.

    You apply a temperature of 110 degrees to B.

    Eventually object B reaches a temperature of 110 degrees because you said: "With no modifications, the heat source is applied to B. Again, all other things being equal: should B be expected to ...".

    The source of heat used to heat A is identical to what was used with object B.

Object B loses it's heat because hot flows to cold

A more easy to visualize example might be two identical hot water bottles with equal volumes of warm water, both at the same temperature. One hot water bottle is given to someone whom is freezing cold, possibly just walked out of a walk-in freezer. The other hot water bottle is given to someone whom is on the beach on a very hot day.

Both persons extract all the available heat from the warm water but the freezing person finds that the bottle is hot while the overheated person finds the warm water somewhat cool.

  • $\begingroup$ I was annoyed at first, thinking to myself that this should have been a comment asking for clarification if the heat source temperature matters, but what you did there is a little funny so you will get no downvotes from me. I was assuming that the heat source is hotter than either A or B by a somewhat significant margin (as in the example of a fire in a room), but I take it that you are assuming a closed system because I did not specify otherwise, and carrying on to the logical conclusion that the heat source itself could only by 110 if that's where equilibrium resulted? $\endgroup$ – Aaron Jul 12 '18 at 21:41
  • $\begingroup$ This type of problem is why I asked my next question for further clarification on engineering.SE instead of physics.SE ;) This could be made into one of those "A guy asked the same question of a physicist, an engineer, and..." type of jokes. Seriously though, since I'm asking from a practical point of view rather than pure physics theory, how would you suggest I edit the question to make it clearer (if at all)? $\endgroup$ – Aaron Jul 12 '18 at 21:44
  • $\begingroup$ Bag of money #1 is 100, bag of money #2 is 200. A rich person like yourself comes along and increases bag #1 by 10 so bag #1 is 110. If you give equal to the 2nd bag it's ahead by 10, or does it look 10 short? Does bag #2 gain, lose or remain the same. Not that the analogy is comparable, except the part about annoyed at first and ahead vs. behind ... $\endgroup$ – Rob Jul 13 '18 at 1:25

The answer to this question is strongly material dependent!

1st law thermodynamics tells us that if we heat up a body by an infinitesimal amount dQ but with no work done and no volume changes on the body, then dQ=dU, where dU stands for a corresponding infinitesimal increase in the internal energy of the body. Now, we said there are no volume changes, we naively expect the internal energy to depend only on temperature. Therefore we can write:

$dQ=\big(\frac{dU}{dT}\big)dT=mC(T)dT$, $\hspace{0.3cm}$ $mC(T):= \frac{dU}{dT}$

giving us the well known equation that describes how heat translates into temperature differences, with a small twist! Here the specific heat can depend on the temperature.

Now in most introductory contexts, it is usually assumed that the specific heat $C(T)$ is a temperature independent constant. In some real materials (e.g. metals at some high temperature, gases close to a classical ideal one) assuming the specific heat is a constant is a great approximation and indeed for the same amount of heat you'll get the same increase in temperature no matter what your initial temperature is, because then it is true that:


and the temperature difference is directly proportional to the heat distributed.

However if the material's specific heat does depend on temperature strongly, ($C$ is not a constant anymore), then this not true anymore. I'll finish with an example. For a cold gas of free electrons, it is well known that $C(T)=\gamma T$, with $\gamma$ being a constant. By integrating the equation above one can see that for a specific amount of heat Q given the following holds:


so for different initial temperatures $T_i$ but the same amount of heat $Q$ you can see that you don't get the same temperature differences.

  • $\begingroup$ What is $a$? Do you mean "$\gamma$ being a constant"? $\endgroup$ – Bill N Jul 12 '18 at 17:37

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