I know that heat is energy in transit. In other words, there is heat transfer because particles (of the hotter object) impart their kinetic energy to other particles (of the colder object) through collisions.

1) So, what is the difference between heat and kinetic energy ?

One particular source says that there is no difference between the two and that "heat is the kinetic energy stored microscopically as random motion of many particles". But another source says that heat cannot be stored. I am completely confused.

2) This is another definition I found:

"heat Q is defined for a given process as the internal energy interchanged which is neither work nor due to flow of matter


Notice that internal energy is a state function and ΔU denotes the difference between the initial and final energies, but heat is not a state" function and this is why we write Q instead of an incorrect ΔQ"

This one seems to have more depth but I just cannot understand what the author means when he says "which is neither work nor due to flow of matter".


Indeed it is not quite correct to say that heat is energy stored microscopically as random motion of many particles. The thermodynamic definition of heat is given in the following way (for systems with fixed number of particles):

We experimentally observe that the amount of work, $W_{\mathrm{ad}}$, needed to bring a thermodynamic state $A$ to a state $B$ is independent of the process as long as the process is adiabatic. This allows us to define the internal energy $U$ as $\Delta U=-W_{\mathrm{ad}}$, where $W_{\mathrm{ad}}$ is the work done by the system on an adiabatic process. This a state function since it does depend only on the initial and final states and it is defined for any states $A$ and $B$. However, if we go from $A$ to $B$ through a non adiabatic process, then in general $\Delta U\neq -W_{\mathrm{ad}}$, there might be some excess or missing of energy. This difference is corrected by adding a new term, $Q$, which correspond to the energy we were not able to account for as work: $\Delta U= Q-W$.

This definition of heat can also be justified from a microscopic point of view. Heat is 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. For a microscopic system it is impossible to take account for every individual work on the molecules so we are forced to interpret all these molecular works as heat.

  • $\begingroup$ Why ΔU=−W(ad) ? Why is the negative sign required ? $\endgroup$ – Gokulakrishnan Shankar Mar 25 '18 at 11:25
  • $\begingroup$ @GokulakrishnanShankar The negative sign is an appropriate convention. It means that if the system does work, then its internal energy decreases. $\endgroup$ – Diracology Mar 25 '18 at 12:43

In a gas, heat is the random kinetic energy of the gas molecules. Within a small region of the gas, that means the kinetic energy of all non-random motion is not considered to be part of the heat energy. So, the energies in flow, rotation, expansion/contraction, etc., need to be subtracted out.

Quite frankly, "random" means something akin to "unknown" or "unknowable" in this context. If the details of the motion of all the molecules are known, then the energy of the motion can be extracted directly (in principle) without using a heat engine. So if flow, rotation, local expansion/contraction, etc., are unknowable they are effectively random, so their energy gets lumped in with heat energy. In a system with just a few molecules, "heat" and "temperature" are very vague concepts.

  • $\begingroup$ But kinetic energy does consist only of flow, rotation, expansion/contraction, etc, so what will be left on subtracting these? $\endgroup$ – Gokulakrishnan Shankar Mar 25 '18 at 11:11
  • $\begingroup$ In classical, non-quantum mechanical world, you would be right. However, it is truly impossible to know every detail of the motion of molecules, atoms, or anything like that. Quantum mechanics prevents us from knowing the position x and momentum p of, for example, an electron to an accuracy any better than Δ x Δ p ≳ h , where Δ x means "uncertainty in x" and Δ p means "uncertainty in y". So on the quantum mechanical scale, we can't know what to subtract. Nature itself wouldn't know what to subtract, because of inherent quantum indeterminacy. $\endgroup$ – S. McGrew Mar 25 '18 at 17:20
  • $\begingroup$ In reality, there irreducible randomness: motions and configurations that we can't even in principle measure and subtract out. $\endgroup$ – S. McGrew Mar 25 '18 at 17:20

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