Question. Why heat is an extensive quantity? That is, why heat is proportional to mass? I am interested in an answer based on classical thermodynamics. I am asking to help me fill the gaps in my argument or to give me better argument.

Weary similar questions answer on which should imply answer to mine question. So if you answer on one of these you answer on my question.

  • Why heat is additive quantity?
  • Why internal energy change $\Delta U$ is extensive quantity?
  • It is possible to formulate these questions like this $m \cdot \left (\frac{\partial U}{\partial m} \right)_{T,p,n}=U$ , $m \cdot\left (\frac{\partial U}{\partial m} \right)_{T,V,n}=U$ or like $m \cdot \left (\frac{\partial Q}{\partial m} \right)_{T,p,n}=Q$, $m \cdot \left (\frac{\partial Q}{\partial m} \right)_{T,V,n}=Q$.

This question has issues with its validity. If we consider heat as function of surface area, then it shouldn't be extensive quantity, because doubling sphere surface mass will increase more than two times and so the heat will do. So I guess we can not call quantity extensive or intensive without mentioning variables upon witch this quantity is dependent and witch are changed. So we can't prove that "heat is extensive" by proving that it is extensive for specific case when only temperature changes. The best way to solve this question I see is by deriving expression of heat change from other quantities for which we can experimentally verify that they are extensive.

Relevance. From heat extensionality follows that entropy is extensional and so is internal energy is extensional and so is enthalpy and so is Gibbs energy. Extensionality of these quantities is used in various proofs. Like this:

  1. Why internal energy $U(S, V, N)$ is a homogeneous function of $S$, $V$, $N$?. Similarly we can prove that every extensive variable that is a function of extensive variables is a homogeneous function of them. Using Euler's homogeneous function theorem we can find integrals of these extensive function differentials.

  2. First proof is used to prove this: Why does $U = T S - P V + \sum_i \mu_i N_i$?

  3. I am grasping that extensionality of Gibbs energy is used to prove Duhem equation.

  4. I am grasping that extensiveness of entropy relies on extensiveness of heat like this: Why is entropy an extensive property?

Probably there are many other useful proofs that rely on extensiveness of heat.

Similar question. Although question is similar, answers seemed insufficient: Is heat an extensive or intensive property?. Here it is told that heat is thermal energy, and energy is extensive so heat is extensive. It sounds right, but I would like to see proof based on laws of physics, definitions and calculus.

My attempt to solve this question. The best way I see to answer this question is to provide equation of $\Delta Q(p,T, n)$ and $\Delta Q(V,T, n)$ where mass is common factor. I can do this by expressing heat as path integral of its differential. It is possible to prove experimentally that corresponding partial derivatives are extensive quantities, we can do it for many substances, and then infer inductively that these quantities are extensive for all materials. Is there batter way?

Gaps in my attempt to prove extensionality of heat.

  • I. So I attempted to do derive these equations, but I am not sure how to account for phase transitions. I am grasping that we can account for phase transitions by using corresponding partial derivatives (for example heat capacity) that contain Dirac delta functions. I guess, other way to do this is to add to the integral heats of phase transitions, but I am not sure how to do this. In case of where heat depends only on temperature we have to add $\sum_i^k \Delta H_{phase}(i \to i)$ in case of constant pressure.
  • II. How can we all be sure that there will be no material where these partial derivatives wont be extensive? If we consider nanoparticles, where surface are is big relative to volume, then there might be problems.
  • III. I have doubts about case where we wary quantity $dn$ of added substance. Chemical potential is intensive quantity. $dn$ Is extensive quantity. So we are adding extensive quantity, we should gain extensive quantity $dQ$. But there is a problem, we consider masses of different substances.

My attempt to solve this question for specific case where only temperature is changed. I am grasping that extensiveness of heat change at constant pressure can be shown like this:

a1. $\Delta Q=\int_{T_1}^{T_2} m C_p dT $ These way we measure heat if there is no phase change.

a3. $\Delta Q=\int_{T_1}^{T_2} dq_{rev}(1->2)+\int_{T_2}^{T_3} dq_{rev}(2->3)+...$ follows from calculus

a3. $\Delta Q=\int_{T_1}^{T_2} m C_p(1->2)dT+\int_{T_2}^{T_3} mC_p(2->3) dT+...$ follovs from a1, a2 using algebra. This if for the case where are no phase transitions.

My attempt to solve this question for case where only the pressure and temperature are being changed.

Similarly in case if volume is constant. I am grasping that similarly, using calculus we can define $\Delta Q$ knowing $C_p(T,P)$.

b1. $dQ=\left (\frac{\partial Q}{\partial T} \right)_{V} dT+\left (\frac{\partial Q}{\partial V} \right)_{T} dV$ so we assume that Q has conditionos partial derivatives of V and T

b2. $dQ=dU+pdV$ so we assume that there is no other work than expansion work.

b3. $m C_V =\left (\frac{\partial Q}{\partial T} \right)_V$ definition

b4. $p(T,V)=\left (\frac{\partial Q}{\partial V} \right)_T$ from b1 and b2

b5. $dQ=m C_V(T,V)dT+p(T,V)dV$ from b1-b4.

b6. $\Delta Q(L)=\int_L \left(m C_V(T,V,1\to2) dT +p(T,V, 1\to2)dV \right)$ From b5 we can infer this for case if there are no phase transition. I am not sure how will this integral look in case if we consider phase transition.

My attempt to solve this question for case where temperature and volume are changed.

c1. $dQ=\left (\frac{\partial Q}{\partial T} \right)_{p} dT+\left (\frac{\partial Q}{\partial p} \right)_{T} dp$

c2. $m C_p=\left (\frac{\partial Q}{\partial T} \right)_{p}$ this is definition.

c3. $dQ=dU+pdV$ first law of thermodynamics, assuming that there is no work other than expansion work.

c4. $pdV=\left (\frac{\partial Q}{\partial p} \right)_T dp$ we obtain this from c1 and c3 using algebra.

$\left (\frac{\partial Q}{\partial p} \right)_T=\frac{dp}{dV} \cdot\frac{1}{p}$ we get it from previous equation using algebra

$\left (\frac{\partial Q}{\partial p} \right)_T=\left(\left(\frac{\partial p}{\partial V}\right)_T+\left(\frac{\partial p}{\partial T} \right)_V \frac{dT}{dV} \right) \cdot\frac{1}{p}$ we gain this by deviding dp(V,T) definition by dV and inserting in previous equation.

$\left (\frac{\partial Q}{\partial p} \right)_T=\left(\frac{\partial p}{\partial V}\right)_T \cdot\frac{1}{p}$ we gain this by setting dT=0

c5. $dQ=mC_p dT+\left (\frac{\partial p}{\partial V} \right)_T \cdot \frac{1}{p}$

c6. $\Delta Q(L)=\int_L \left(m C_p(T,p,1\to2) dT +\left (\frac{\partial p}{\partial V} \right)_T \cdot \frac{1}{p}dp \right)$ It is for the case where is no phase change.

c7. So we should obtain equation where mass is common factor, so $\Delta Q$ is extensive. $\left (\frac{\partial p}{\partial V} \right)_T \cdot \frac{1}{p}$ should be extensive, because it is equal to $pdV=Vp-Vdp$, witch is extensive.

In cases where temperature, volume and added reagent amounts are changed.

d1. $dQ=\left (\frac{\partial Q}{\partial T} \right)_{V} dT+\left (\frac{\partial Q}{\partial V} \right)_{T} dV+\left( \frac{\partial Q}{\partial n}\right )_{T,V} dn$

d2. $dQ=dU+pdV+\mu dn$ I am grasping that chemical potential is a work dont to add new compound to the mixture. Is it so?

d3. $dQ=m C_V dT+pdV+\mu dn$ we do this similarly as in examples above.

d4. So we integrate as previous example. It is for the case where there are no phase change.

d5. So we will obtain equation where mass is common factor, so $\Delta Q$ is extensive.

In cases where temperature, pressure and added reagent amounts are changed.

e1. $dQ=\left (\frac{\partial Q}{\partial T} \right)_{p} dT+\left (\frac{\partial Q}{\partial p} \right)_{T} dp+\left( \frac{\partial Q}{\partial n}\right )_{T,p} dn$ from deffinition of total differential.

e2. $dQ=m C_p dT+pdV+\mu dn$ I am grasping that chemical potential is work from addition of other component to the mixture.

e3. $dQ=m C_p dT+\left(\frac{\partial p}{\partial V}\right)_T \cdot\frac{1}{p} dp+\mu dn$ we do this similarly as in examples above

e4. So we integrate this. This is for the case where is no phase change.

e5. So we will obtain equation where mass is common factor, so $\Delta Q$ is extensive.

P.S.: I am chemist, so things that are obvious to physicists might not be obvious to me. So I prefer proofs. Proof is a sequence of formulas where each of them is an axiom or hypothesis, or derived from previous steps by inference rules. I am better with Fitch notation. It is very good if the proof comes from a book or publication.

Update. First, when I asked my question, question was:"why heat is extensive property?". I din't know that we can't designate functionals as "properties", so I changed word to "quantities". It turned out that there is no "extensive quantity" concept. So right question is: "why heat is proportional to mass?", but I didn't change name of the question because It will become hard to understand some answers.

  • 1
    $\begingroup$ It seems to me that the question is ambiguous: heat is not a "property". It is a quantity of energy exchanged. And so talking about extensivity of heat is problematic. I think what you mean is that, for the same transformation, the amount of heat exchanged is doubled if we double the quantities of matter in the system? $\endgroup$ Commented Dec 5, 2021 at 11:16
  • 1
    $\begingroup$ To complete my comment: a derivative like $\frac{ \partial Q} {\partial T} $ has no meaning (in general) in thermodynamics because there is no state function $Q$ $\endgroup$ Commented Dec 5, 2021 at 12:36
  • $\begingroup$ What makes me perplexed is your reference to being a chemist. I actually got the impression of you having a mathematical attitude. I would never put efforts on reading or writing so much relations when it is clear to me that in general heat is not a property of a system. $\endgroup$
    – Alchimista
    Commented Dec 5, 2021 at 14:47
  • $\begingroup$ Sorry. English is not my native language. Until now I assumed that words "quantity" and "property" are synonyms. $\endgroup$
    – Alex Alex
    Commented Dec 5, 2021 at 15:05
  • $\begingroup$ What is your definition of "extensive". In thermodynamics the terms "extensive" and "intensive" normally refer to thermodynamic properties which heat is not $\endgroup$
    – Bob D
    Commented Dec 5, 2021 at 15:29

3 Answers 3


Physicist answer:

Heat is neither intensive or extensive. Heat is not a property of a system, energy is.

According to 1. law of thermodynamics $\Delta U = Q + W$, the change in energy for a system is the added heat and work.

If no heat or work is added, then the energy is conserved. No heat don't mean that the system is without energy. Conceptionally both work and heat are external concepts - something that is added (or subtracted) to/from the system.

When you double the sphere surface mass with an identical mass you add both particles N and Heat Q to the total system. The result is a system with more particles but not more heat - instead it has higher energy.

  • $\begingroup$ Could you provide reference to the book or publication where is written, that heat is extensive by definition? As I have sad that there are issue with "heat is extensive" if we chose wrong variables. Suppose I want to use other variables than p, V, n, how can I tell that heat will continue to be extensive property when these variables are changed? $\endgroup$
    – Alex Alex
    Commented Dec 5, 2021 at 11:07
  • 3
    $\begingroup$ Quoting "An introduction to Thermal Physics" Daniel V. Schrödinger: "Heat is defined as any spontaneous flow of energy from one object to another, caused by a difference in temperature between the two objects". $\endgroup$
    – Karsten B
    Commented Dec 5, 2021 at 11:18
  • $\begingroup$ I can't infer from that quote that heat is extensive. I don't see the connection between these two statements. Can you provide more explicit citate or explain the connection between it and my question. $\endgroup$
    – Alex Alex
    Commented Dec 5, 2021 at 11:23
  • $\begingroup$ Some thing are self-evident for physicists. I believe that they are self-evident because physicists work weary frequently with them. It goes from physiology, if you do operations weary frequently they become automatic, and you stop noticing the details. It is like writing with a pen, you don't think how you do it. Physicist and chemist have different basic skill sets. $\endgroup$
    – Alex Alex
    Commented Dec 5, 2021 at 11:47
  • 1
    $\begingroup$ @AlexAlex A liquid that evaporates cools down the system. You are confusing the T of the liquid with the T of system / surroundings just to rebut the clear answer by Karsten B. Of course the evaporation of 2 kg water absorbs the double amount of heat as if it would be 1 kg. But it does not makes heat a property of water, and as such there is no reason for the question. $\endgroup$
    – Alchimista
    Commented Dec 5, 2021 at 14:52

Heat is one form of energy exchanged between a system and its surroundings during a process. Or, by the IUPAC definition, heat is energy transferred from a hot to a cold body due to a temperature gradient.

Because heat transfers between two objects (system - surroundings or hot object - cold object), we cannot assign heat unequivocally as an inherent property for either object.

The terms extensive and intensive are applied to distinguish whether an inherent property for a specific system or object is being referenced to the amount (volume, mass, moles) for that specific system or object. We trip into a fallacy when we attempt to apply the terms extensive or intensive to a value that cannot unequivocally be assigned as an inherent property for a specific system or object.

In summary, we cannot define heat as being extensive. Conversely and perhaps more clearly, we cannot define heat as being inherently not intensive.

We may choose to normalize the amount (quantity or value) of heat transferred during a process of energy exchange. We can normalize by time (heat flow rate), area and time (heat flux), or amount (volume, mass specific or molar specific). Normalizing heat quantity by an amount of the system or object does not change the heat quantity into an intensive property or intrinsic quantity for the system or object. Doing so only normalizes to a chosen reference system or object.

In summary, since we can choose any reference system or object to normalize heat transferred during a process, we again demonstrate that heat can never be classed an inherent property, characteristic, quantity, or value for one specific system or object. We again demonstrate that heat cannot be classed as an extensive or intensive property, value, character, or quantity.

  • $\begingroup$ There is a gap in your definition of heat. Heat can flow when there is no temperature gradient, for example in phase transforms. If you are heating water that is at 100C, there is no temperature gradient, but there is heat flow. There is isothermal expansion of gas, and there is also heat flow. For example, Carnot engine gains and loses heat isothermaly. $\endgroup$
    – Alex Alex
    Commented Dec 5, 2021 at 16:56
  • 1
    $\begingroup$ The term isothermal means only that the temperature in the system or object remains fixed during the process that heat is transferred out or into it. The system and surroundings must have different temperatures when heat flows because heat can only flow due to a temperature gradient or difference. This is the IUPAC definition. Take your complaint to them when you feel that you have unequivocally rationalized a true counter example. The idealized isothermal Carnot step with $T_{sys} = T_{surr}$ during the heat flow is a hypothetical case that will never exist in reality. $\endgroup$ Commented Dec 5, 2021 at 17:17
  • $\begingroup$ @Alchimista - Your edits are appreciated. It seems that I had semiconductors on my mind. $\endgroup$ Commented Dec 7, 2021 at 22:33

Maybe it is useful to use the microscopic picture of the problem. Say that you have a system with $N$ identical non-interacting particles, the spectrum of individual energies is discrete so the total energy of the system is:

$U=\sum_{i=1}^{N} \epsilon_i$

If the number of particles is high, you can rewrite the above sum as:


Where $<\epsilon>$ is the average of the individual energies in the system. We could compute this average with the statistical mechanics' ensemble theory, but this is not the subject we are interested in. The above equation just implies that the internal energy is extensive.

The effect of heat transfer in the system can be seen as a change in the occupancy levels of particles. Due to the first law, the change in internal energy is precisely the heat transfer (we suppose $W=0$):

$\Delta U=Q=\sum_{i=0}^{N}(\epsilon_i -\epsilon'_i)$.

Where $\{\epsilon'_i\}_{i=1,...,N}$ are the new energies of the particles once heat transfer affected the system. To me, the above equation is stating that $Q$ does not have to be fundamentally proportional to $N$ (this is, $Q$ is not, in principle, extensive). For example, if the heat transfer is so small that it only affects one particle (say particle labeled $j$).

$Q=\Delta U=\epsilon_j-\epsilon'_j$

Which is not proportional to $N$. Interestingly, this change of energy would be negligible in the thermodynamic limit. Using the previous example:

$U'=U+\Delta U=N<\epsilon>+\epsilon_j-\epsilon'_j \approx N<\epsilon>$

Heat transfer is non-negligible when it affects on average to all particles of the system. In this case, $Q$ is proportional to the number of particles.

$Q=N(<\epsilon>-<\epsilon'>)\rightarrow U'=N<\epsilon'>$.

To summarize, $Q$ does not have to be fundamentally proportional to the number of particles. However, $Q$ is non-negligible in the thermodynamic limit when is proportional to $N$.

  • $\begingroup$ Thank you. As I have sad in the begging of my question, I am interested only in answer based on classical thermodynamics. Why so? Because of Occam razor, I want infer this answer from smallest possible and simplest possible set premises. As I have sad in P.S. I am chemist, so I have different set of primitive concepts. Your answer is incomprehensible for me. $\endgroup$
    – Alex Alex
    Commented Jan 2, 2022 at 11:05
  • $\begingroup$ My bad. All I was trying to say is that the statement "$Q$ is extensive" is an assumption rather than a deduction. The internal energy has to be extensive, whereas heat is not inherently extensive. You can heat a glass of water introducing an amount of energy that is not proportional to the number of molecules of water! However, when the system is macroscopic, heat induces non-negligible changes in the internal energy only when is extensive. $\endgroup$
    – Javi
    Commented Jan 2, 2022 at 17:22

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