The temperature change in an isothermal process is zero. As a result (if the system is made of an ideal gas) the change in internal energy must also be zero. I don't want to consult the equations. I'm trying to understand this with logic.

Temperature change accounts for heat flow. If there's no change in temperature how can we say the heat supplied to or the heat released from the system is non zero?

My second question : Let's consider a simple example of state change (like the melting of ice) at constant temperature, heat is supplied to the system and the system is in thermal equilibrium, how is work being done here?

  • $\begingroup$ One place the energy can go if you add heat to a gas is toward a temperature increase. Another, if you let the gas expand, is toward work on the environment. $\endgroup$
    – Ben51
    Jan 12, 2018 at 2:52
  • $\begingroup$ In an isothermal process, the system is in thermal equilibrium with its surroundings. For thermal equilibrium with its surroundings, the walls of the system must be diathermal i.e. conducting walls. Consider an adiabatic expansion of an ideal gas. As the walls are adiabatic i.e. no heat exchange takes place, the temperature of the system decreases. $\endgroup$ Jan 12, 2018 at 2:52
  • $\begingroup$ How is there a change in heat energy ? Say i had x joules of heat energy when the system is at a constant temperature, and the heat supplied will be used to do work on the surroundings and thus the heat energy supplied to the surrounding will be x+y joules and thus my change will be y joules? $\endgroup$
    – susan J
    Jan 12, 2018 at 2:54
  • $\begingroup$ youtu.be/4i1MUWJoI0U $\endgroup$ Jan 12, 2018 at 2:57
  • $\begingroup$ so is my above assertion on y joules of heat energy correct?? $\endgroup$
    – susan J
    Jan 12, 2018 at 3:00

2 Answers 2


You have misunderstood a subtle connection in thermodynamics. Heat flow does not necessarily mean temperature must change. I think you are also missing a key concept in the understanding of the term "heat".

When studying thermodynamics, it is much more accurate (and also much more helpful, I think) to consider heat simply as "energy that is transferred from one object to another due to a difference in temperature". Now that last part is very important; heat is not some magical or special form of energy, it is energy. The only special part is that it is mainly connected to temperature and occurs only due to a difference in temperature.

In fact, due to its definition, it is also somewhat inaccurate to say that an object has "heat" per se. An object has thermal energy, yes, but not exactly "heat" because "heat" is only transferred, not contained (although many people and physicists use it that way all the time because what they actually mean is well understood). If you really want to delve deeper into heat, see this excellent answer by Mark. The wikipedia page on heat is also very helpful.

Think of it like this; say I have a trapped gas in a closed piston with a very smooth piston walls so it has almost no friction. If I attach a small weight to that piston and slowly pump heat into my system and let that system attain equilibrium for each small step of the way, that gas might expand and do work against/on the small weight that piston-rod is connected to instead of increasing the temperature of the system. Here heat is being transformed very obviously into work because you see the weight moving, even if it's slow.

Your ice to water example is a different case but a good one when trying to probe into what that heat is doing. Indeed, it goes back to the very heart of understanding heat and energy. Heat, in this case of phase change (or state change) of water, is being channeled into overcoming the intermolecular forces of ice. In other words, the heat is being used to break the crystal structure of ice rather than to increase the average kinetic energy of the molecules. The heat is doing work, just not macroscopic work.

  • $\begingroup$ So, if I think of an adiabatic system as the one where no heat is possessed by the system , Q = 0 and, that the energy possessed by the system will only be it's internal energy and the work done on / by the system is because of internal energy . Is my thought process correct? $\endgroup$
    – susan J
    Jan 12, 2018 at 4:09
  • 1
    $\begingroup$ Actually, disregard my last comment. Yes, you are mostly correct. However, keep in mind that an adiabatic system is not a system that possesses no heat but rather a system where no heat can enter or leave. For most formulations of the First Law, setting it to be $Q=0$ is acceptable, as Q is heat and heat must necessarily flow. Me pointing out that it should be $\Delta Q=0$ rather than $Q=0$ is just me saying that it is the heat going in and out that is zero, not the "amount" of heat. Using $Q=0$ is fine. $\endgroup$ Jan 12, 2018 at 4:25
  • $\begingroup$ @susanJ Just for the record, heat is not something we can "possess". Heat is a transfer of energy only. Just like work. We cannot "possess" work but we can do/create work. $\endgroup$
    – Steeven
    Jan 12, 2018 at 9:11

If there's no change in temperature how can we say the heat supplied to or the heat released from the system is non zero?

Heat and temperature are not the same thing. Heat is a mode of energy transfer whereas the latter is a measure of average kinetic energy. Heat flow may or may not change the temperature. If the heat add in to the internal kinetic energy of the system you will observe a change in temperature but if something else happens while, heat is being transferred, which robs the system of its internal kinetic energy, you will observe no change in temperature.

For instance, consider a closed system with moveable piston. If there is no net change in temperature and you wish to find about heat transfer to (or from) the system, you have to look at the work done by (or on) system. According to the first law of thermodynamics,

$$\Delta U=q-P\Delta V$$

For an isothermal process,

$q=P\Delta V$

We have a given condition that the temperature of the system is constant. So if something happens in the system which increases its $U$, something else must happen which could nullify this increment, so that the net effect of such a process brings no change in the internal energy of the system.

Heat supplied or work done on the system increases the internal energy and heat released or work done by the system decreases the internal energy. So if heat is supplied to the system, work has to be done $by$ the system, if the temperature has to remain constant. And these two energy terms operate opposite to each other.

State change (like the melting of ice )at constant temperature, heat is supplied to the system and the system is in thermal equilibrium, how is work being done here?

$\Delta U$ is zero at constant temperature only for ideal gases. Internal energy contains both kinetic energy and potential energy terms, but since ideal gases do not interact with each other, the potential energy term is removed.

However, in the real world potential energy comes into play. The heat supplied externally adds in to the internal potential energy of ice. No change occurs in the internal kinetic energy of the system, which implies that the temperature remains constant $(\Delta T=0)$.

  • $\begingroup$ A different question kinda not related to your answer but,In thermodynamics Work is given by PdeltaV. Why isn't it delta P x delta V doesn't pressure change as the gases expand or compress? $\endgroup$
    – susan J
    Jan 12, 2018 at 7:07
  • $\begingroup$ It definitely can. But if you consider external pressure which remains constant for a process, you wont need to consider internal pressure, as work done by external agent is equal to work done by the system. There are other cases, like reversible processes, where you have to establish a connection between the pressure inside the system with the volume using the ideal gas equation, which presents you with a integral which can give you the work. physics.stackexchange.com/questions/369188/… $\endgroup$
    – Mitchell
    Jan 12, 2018 at 7:16

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