# Relationship between heat and temperature

I'm reading a book called "A Guide to Thermal Physics" by Chris McMullen and I came across a passage that does not entirely sit well with me.

" It may seem very intuitive to want to associate absorption of heat with an increase in temperature and heat loss with a decrease in temperature, as suggested by the equation for heat capacity. However, the heat that an object exchanges with its surroundings does not necessarily go toward changing the temperature of the object – it is also related to the work done and to latent heat of transformation. For example, a system may absorb heat and use this thermal energy exclusively to do work, without changing temperature, or a substance may absorb heat and use this thermal energy to undergo a phase transition rather than change temperature. The important point is that a process that is isothermal may not be adiabatic, and a process that is adiabatic may not be isothermal. That is, a system can absorb heat without changing temperature, and a system can change temperature without exchanging any heat."

How can a system change temperature without exchanging heat? Also, he says a system can use thermal energy to do work - I thought that thermal energy was essentially that energy which cannot be used for work because it's just the random motion of molecules which can not be "recaptured" the way mechanical energy like pushing a piston can be reversed.

How can a system change temperature without exchanging heat?

That is what an adiabatic process means! Have an insulated gas piston, i.e. never exchanging heat, and you push on it. The work done you pushing on the piston will be converted into temperature change

Also, he says a system can use thermal energy to do work - I thought that thermal energy was essentially that energy which cannot be used for work because it's just the random motion of molecules which can not be "recaptured" the way mechanical energy like pushing a piston can be reversed.

The quote is referring to a heat engine. A heat engine takes in thermal energy (= heat) at high temperature, and expels a little less of that heat at low temperature, and uses the difference to do work.

If you only have one temperature to work with, then, yes, you cannot recapture thermal energy to do work. But you can, if you have two different temperature heat baths to work in-between.

• Not to be pedantic, but the work you did to push the piston is essentially transferring energy to the system, right? So how is that transfer of energy via work different than heat? Is it just a conceptual distinction we make to be keep terminology clear? Also, when you say "one temperature to work with" do you mean like say a cup of hot coffee that is gradually cooling due to air exposure, so you can't recapture that thermal heat that is going to the air? Commented Dec 15, 2023 at 8:42
• "Is it just a conceptual distinction we make to keep terminology clear?" Half and half. Heat flow is a transfer of energy that is automatic, statistical, only dependent upon temperature differences and we can only control it by changing the wall material. Work, on the other hand, is ordered, easily convertible and controllable, and so forth. Also, before thermodynamics became a proper theory, people did not know that heat is a form of energy at all. The first law is essentially stating that energy is conserved, and if you think you missed some, that missed part is heat. Commented Dec 15, 2023 at 9:44
• "you can't recapture that thermal heat that is going to the air" NO! That is precisely the opposite of what I meant. Hot coffee is at a different temperature than air, and so the difference in temperature allows us to extract work, if we run a heat engine between them. That is how engines running on a hot cup works. One single temperature, means one single temperature. A room in which everything is already at one single temperature. Then the heat in it cannot be used. Commented Dec 15, 2023 at 9:47
• You can recapture thermal energy to do work. All you need to do is allow the gas to expand against a slightly lower pressure. You just can't do work using a single thermal reservoir cyclically. Commented Dec 15, 2023 at 12:17

Have you ever pumped air into your bicycle tyres using manual pump? When you push the piston in surface of pump gets warmer and when you pull it out it gets colder. $$dU = -PdV + TdS$$ $$dU + PdV = TdS$$ $$dQ = dU + PdV$$ $$dQ = TdS$$ $$T = \frac{dQ}{dS}$$ At constant volume $$T = \frac{dU}{dS}$$

Hence Temperature can change by heat, change in Entropy or internal energy.

However,

a system may absorb heat and use this thermal energy exclusively to do work, without changing temperature.

This statement has to be taken with grain of salt. Heat is low grade energy and cannot be fully converted to work. This is gist of second law of thermodynamics.

Whereas Helmholtz Free Energy is a thermodynamic potential that is equal to reversible work that can be obtained from system for isothermal change.

While dealing with thermodynamics can be tricky I will suggest you to try understanding it mathematically along with intuition.

In words (no equations):

• Temperature is a measure of the amount of energy contained in matter.

• The amount of energy contained in matter can be changed via heat, work or both.

Therefore, if we add energy in the form of work the internal energy increases, and so does temperature, even though there is no heat involved.

A simple demonstration: Rub your hands, do they feel warmer? The work done by rubbing adds energy and increases the temperature of the hands.

The opposite is also true: in adiabatic expansion a gas cools. Adiabatic of course means no exchange of heat.

How can a system change temperature without exchanging heat?

A gas in a thermally isolated container will expand and cool down if the container is expanded. For air this temperature drop should be 3 degrees if the volume is doubled (see this page) https://en.wikipedia.org/wiki/Joule_expansion

More generally, a system with constant energy $$E = T + V$$ that has the option to reduce its potential energy $$V-\delta E$$ will increase its kinetic energy $$T + \delta E$$, and thus increase its temperature.

Take a fluid in a thermally isolated container prepared in a non-equilibrium state : This could be a binary mixture of two fluid components in the mixed state, or a binary electrolyte with a density gradient created by an external electric field.

Without external constrains these systems spontaneously decay to their equilibrium structures : The binary mixture (such as oil and water) phase separates, density gradients in the electrolyte (such as salt in water) are removed, and in all cases low energy structures are achieved and so there is a release of energy. So where should that energy go ? If heat exchange is forbidden by thermal isolation, then the temperature will rise.

In computational simulations of thermodynamic systems, these kinds of temperature drifts due to dissipative structures are commonly encountered. This is why simulations of thermodynamic systems need to have additional constrains for fixing the temperature in addition to the underlying microscopic Newtons equations.