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In textbooks, heat is usually defined as the energy transfer due to temperature difference.

However, we don't know what temperature is in the first place. I think it's better to define heat first and then define temperature $1/T$ as the integrating factor so that one has $$\oint\frac{\delta Q}{T}=0$$ and hence define entropy as the potential $$\Delta S=\int\frac{\delta Q}{T}$$

So my problem is I cannot use temperature or entropy when defining heat.

So what is the definition of heat?

I think my problem can be solved if one can define either heat or work without mentioning temperature and entropy. For instance, if we can somehow define work, then heat can be define as the energy change not in the form of work. And vice versa.

So in summary, my question is how to define work or heat without mentioning temperature and entropy in thermodynamics (without referring to statistical mechanics)?

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    $\begingroup$ The ultimate problem, in my opinion, is that you can't define energy in more fundamental terms, and that every other definition refers back to energy. $\endgroup$ – user108787 Nov 8 '16 at 0:38
  • $\begingroup$ I am comfortable to take energy for granted. But in thermodynamics it is said that there are two different ways of energy transfer. So what in fact is the difference, without mentioning temperature and entropy? $\endgroup$ – velut luna Nov 8 '16 at 0:43
  • $\begingroup$ Honestly, I don't know......I have recently gone back to thermodynamics as a break from quantum theory and because I was not aware of how much physics thermodynamics incorporates......ask me in 4 months : ) best of luck with your post. $\endgroup$ – user108787 Nov 8 '16 at 0:50
  • $\begingroup$ With the so-called zeroth law one can define temperature unambiguously without reference to internal energy or anything else not already present in mechanics.With the first law stating that internal energy is conserved in all interactions and postulating the existence of adiabatic interaction (pure work) in which energy is conserved one gets "heat" as the deficit between internal energy and work in diabatic interactions. $\endgroup$ – hyportnex Nov 8 '16 at 1:19
  • $\begingroup$ In thermodynamics, heat and work are not separate.The system will not store them in separate ways, but as its internal energy. As @ CountTo10 said, both are forms of energy. Defining energy beyond that....I don't know $\endgroup$ – UKH Nov 8 '16 at 1:35
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Essentially, Callen gave the following description in the first chapter of his book, "An Introduction To Thermostatics":

There are say $10^{23}$ atoms in a system and each atom can be described by 3 coordinates. Now out of all possible combinations of the motion(also called modes) of these atoms, only a few particular combinations of atomic coordinates that are essentially time-independent are macroscopically observable. For example, a combination can be in which all the atoms are moving in one direction. Then the system as a whole moves in that direction and that is macroscopically observable. Or perhaps, the atoms are moving away from each other such that the volume of the system increases.

Thus macroscopic observations sense only coarse spatial averages of atomic coordinates. He further says that thermodynamics describes only static states of macroscopic systems.

Finally, energy transferred via a "mechanical mode"(mechanical macroscopic coordinate like volume, stress, etc.) is called mechanical work.

Whereas, energy transfer via "hidden atomic modes" is called "heat".

Hope this helps. Refer to Callen for details. This topic is discussed in the beginning of the first chapter. Here is a link to an online version of the book:

http://keszei.chem.elte.hu/1alapFizkem/H.B.Callen-Thermodynamics.pdf

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I think the best definition of Heat one can give, is, as you mentioned, the Internal energy change that is not due to Work.

For a system, the total energy $E$ is the sum of $E=E_{potential}+E_{kinetic}+U$, where $U$ is the internal energy, that takes into account all the degrees of freedom inside the system. The First Law of thermodynamics states $\Delta U = W + Q$ where W is the work of external forces on the system, and Q is the exchanged Heat.

I think Callen has a good discussion of the topic. Callen, Herbert B. Thermodynamics & an Intro. to Thermostatistics. John wiley & sons, 2006.

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The concept of work in thermodynamics is pretty much the same as the one used in mechanics. Roughly speaking work is motion against forces. Whatever the system is, if it lifts a weight in gravitational field it does work. So work can be provided by rope and pulleys, electric currents, gases and movable pistons, etc.

In thermodynamics one can define heat by following the historical construction of the first law presented by Joule. The (adiabatic) work done on a system thermally isolated is independent of the process so it is convenient to define a state function $U$ (internal energy) which associates a value to every possible state of the system and to define $W=-\Delta U$. If then one removes the thermal isolation the necessary work done on the system to carry it from state $A$ to state $B$ is process dependent. The previous relation between $U$ and $W$ no longer holds but it can be corrected by adding a new term, $\Delta U=Q-W$. This is the thermodynamic definition of the heat $Q$.

Note that this definition of heat does not require the definition of temperature (which is precisely given by the zeroth law) and in fact it can be used to define the calorimetric concept of temperature. For instance through the relations such as $Q=mc\Delta T$.

From a microscopic point of view one can understand heat as any work the molecules of the neighborhood do on the molecules of the system that cannot (in the sense of practical purposes) be expressed by force times distance. Obviously for a microscopic system it is impossible to take account for every individual work on the molecules so we are forced to interpret this molecular work as heat.

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We all know there are four laws in thermodynamics,the Zeroth Law (If system A is in equilibrium with system B and with system C then system B is in equilibrium with system C.) tells us what the temperature is;the Second Law (Heat flows from hot to cold,or It is not possible to convert all heat energy to work.) tells us the what the entropy is and also tell us how to order the temperature at the same time.

If we consider these definitions in the context of statistical mechanics(which is the microscopic version of thermodynamics),the relation between temperature and entropy will be more clear and precise.Actually you will see the entropy is related closely with multiplicity function (the number of microstates $\Omega$ of your macroscopic system) by the following formula . \begin{equation} S = k_B \ln \Omega. \end{equation} If you want to find the equilibrium state(which will be described by some macroscopic physical quantities,such as energy,pressure,volume,temperature $\cdots$ ) then you must find the the maximum of multiplicity function (or just maximize the entropy),finally the extremum point will exactly be defined as the inverse temperature.

As for the concepts of work and heat,we will see them in the First Law of thermodynamics: \begin{equation} \Delta U = T d S + P d V. \end{equation} which is just a expression of energy conservation,namely one can get energy increased by heated or work done. But in statistical mechanics we will see another explanation elegantly.The total internal energy will be written as \begin{equation} U = \sum_i P_i E_i, \end{equation} where i labels some state and $P_i$ and $E_i$ are the probability and energy level in which particle may live,respectively,then \begin{equation} d U = \sum_i E_i d P_i + \sum_i P_i d E_i. \end{equation} So you will find the work done corresponding to the shift of energy levels and the heated corresponding to the change of the probability.

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protected by Qmechanic Nov 25 '16 at 15:09

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