Heat vs Internal Energy: The Difference

The distinction, stated once

One of these is a state, the other is a transfer, and no amount of care with joules will fix a sentence that mixes them up.

Internal energy is a property of the system as it is now. Essential knowledge 9.4.A.1 says the internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects. Freeze a gas in time and ask what its internal energy is, and there is an answer: for an ideal monatomic gas, 9.4.A.1.ii gives it exactly as $U = \frac{3}{2}nRT = \frac{3}{2}Nk_BT$.

Heat is a transfer that happens to the system during an interval. The AP Physics 2 equation sheet's Thermal Physics symbol list defines $Q$ as the energy transferred to a system by heating, and 9.3.A.1.i defines heating as the transfer of energy into a system by thermal processes. Freeze the gas in time and ask what its heat is, and there is no answer, because heat is not a thing a frozen instant can contain.

The CED puts it more bluntly than most textbooks dare. Its appendix, in the section headed Heat vs. Heating vs. Cooling, says that using the physics definition of heat, an object cannot have heat any more than an object may have work.

So the test is grammatical before it is physical:

• "The gas has $8725\ \text{J}$ of internal energy." Correct, and computable.
• "The gas has $8725\ \text{J}$ of heat." Not a claim that can be true or false. It is malformed, in the same way that "the gas has $500\ \text{J}$ of work" is malformed.
• "The gas absorbed $750\ \text{J}$ by heating during that expansion." Correct, because it names an interval.

Heat vs work covers the other half of this: heat and work are two transfers that differ only in mechanism. This page is about the difference between a transfer and the thing it changes.

The sheet writes one delta and not the other

The distinction is printed on your equation sheet, in the notation, and almost nobody notices. The AP Physics 2 Thermal Physics panel gives the first law as

Look at the deltas. There is one, and it is on $U$. Not on $Q$, not on $W$.

That asymmetry is not a stylistic choice. $U$ is a quantity the system possesses, so it has a value before the process and a different value after, and the thing that appears in an equation about a process is the change. $Q$ and $W$ are the process. They have no before value and no after value to subtract, so there is nothing for a delta to operate on. You will never see $\Delta Q$ on an AP equation sheet, and if you write one in a solution you have written something that does not mean anything.

The same panel's symbol list makes the parallel explicit. It defines $U$ as internal energy, full stop, a property. It defines $Q$ as energy transferred to a system by heating and $W$ as work done on a system, both of them phrased as actions with a direction across a boundary.

Two consequences you can use immediately.

First, internal energy has an absolute value and heat does not. $U = \frac{3}{2}nRT$ is a number you can compute from the current state with no history at all. There is no corresponding formula for $Q$ and there cannot be, because until a process is named there is nothing to compute.

Second, $U$ is path independent and $Q$ is not. Two gases at the same $n$ and $T$ have the same internal energy no matter how they got there. Two gases that underwent different processes to reach that state absorbed different amounts of heat. This is why 9.4.B.1 can describe the first law as a restatement of conservation of energy: the left side tracks a bank balance, the right side tracks the deposits and withdrawals.

Heat against internal energy, row by row

Row three is the whole page. Everything else follows from it.

Row six is the one that decides exam questions. Because $U$ has an absolute value, a question can hand you $n$ and $T$ and expect a number. Because $Q$ does not, a question asking for the heat must specify what happened, and if it has not specified, you are being asked for something else.

Row nine is worth a caution, because the asymmetry is easy to misread. Internal energy being positive does not make it more real than heat; it reflects that $\frac{3}{2}nRT$ is positive whenever $T$ is, and that the AP model of an ideal gas has no internal potential energy to be negative. The sign on $Q$ is a direction, not a magnitude comparison.

Row twelve comes from 9.4.A.2, which says changes to a system's internal energy can result in changes to the internal structure and internal behavior of that system without changing the motion of the system's center of mass. The word doing the work in that sentence is internal. Bulk motion of the whole system is a separate account.

The case that separates them: a gas that absorbs heat and gets colder

If heat were a substance the gas stores, then putting heat in would always mean having more of it, and having more would always mean being hotter. Here is a process where the gas absorbs a substantial amount of energy by heating and ends up colder than it started.

Take $0.25\ \text{mol}$ of a monatomic ideal gas and expand it along a straight line on a PV diagram from state A at $P = 3.0 \times 10^5\ \text{Pa}$, $V = 2.0 \times 10^{-3}\ \text{m}^3$ to state B at $P = 0.50 \times 10^5\ \text{Pa}$, $V = 8.0 \times 10^{-3}\ \text{m}^3$.

The state change. $PV$ at A is $600\ \text{J}$ and at B is $400\ \text{J}$, so by $U = \frac{3}{2}nRT$ together with $PV = nRT$, the internal energy is $\frac{3}{2}PV$ and

The internal energy fell. Since $U$ tracks temperature for an ideal gas, the gas is colder at B: $288.8\ \text{K}$ down to $192.5\ \text{K}$.

The transfers. The gas expanded, so it did work on its surroundings and $W$ is negative. Its magnitude is the area under the straight line, which splits into a rectangle of $(0.50 \times 10^5)(6.0 \times 10^{-3}) = 300\ \text{J}$ and a triangle of $\frac{1}{2}(2.5 \times 10^5)(6.0 \times 10^{-3}) = 750\ \text{J}$, giving $1050\ \text{J}$ in total. So $W = -1050\ \text{J}$.

Now the first law, rearranged: $Q = \Delta U - W = -300 - (-1050) = +750\ \text{J}$.

Read that pair of numbers together. The gas took in $750\ \text{J}$ by heating and its internal energy went down by $300\ \text{J}$. Where did it go? Out again as work: $1050\ \text{J}$ left through the piston, $750\ \text{J}$ came in through the walls, and the $300\ \text{J}$ shortfall was made up from the gas's own store.

A system that contained heat could not do this. If the $750\ \text{J}$ that entered were an amount of heat now held by the gas, the gas would have more of it than before, and it does not have any. What it has is internal energy, and it has less. The transfer and the state moved in opposite directions during the same process, which is the cleanest possible proof that they are different quantities.

The reverse is available too. Compress a gas adiabatically and $Q = 0$ throughout while the internal energy climbs steadily. Zero transfer, rising state.

What internal energy is actually made of

Heat has no composition. It is a flow, and asking what it is made of is like asking what a bank transfer is made of. Internal energy does have a composition, and the CED spells it out.

9.4.A.1 gives the general form: the internal energy of a system is the sum of the kinetic energy of the objects that make up the system and the potential energy of the configuration of those objects. Two terms, one for motion and one for arrangement.

For an ideal gas, the second term vanishes. 9.4.A.1.i explains why: the atoms in an ideal gas do not interact with each other via conservative forces, and the internal structure is not considered, therefore an ideal gas does not have internal potential energy. That leaves 9.4.A.1.ii, the internal energy of an ideal monatomic gas is the sum of the kinetic energies of the constituent atoms, with the relevant equation

The exam conventions box on the AP Physics 2 sheet states that ideal gases are monatomic, so that equation is the one in play for this course.

Notice what this buys. Internal energy is written entirely in terms of the current state, $n$ and $T$. No time, no history, no process. It is a state function by construction.

Now the boundary of the word internal, which is 9.4.A.2: changes to a system's internal energy can result in changes to the internal structure and internal behavior of that system without changing the motion of the system's center of mass. Read it in both directions. A gas can heat up without going anywhere, and a gas can go somewhere without heating up.

So the kinetic energy of a sealed cylinder riding on a truck at $20\ \text{m/s}$ is not part of its internal energy. Those atoms are all drifting along together, and the internal account only counts their motion relative to the center of mass. Stop the truck and the internal energy is unchanged, unless something actually transfers energy into the gas. This matters more than it sounds: it is the reason $\Delta U = Q + W$ can be applied to a gas in a moving vehicle without carrying a bulk kinetic energy term, and it is the reason temperature is not a measure of how fast something is travelling.

Heat, by contrast, has no inside and no outside of its own. It is defined only by the boundary it crosses, which is why every statement about $Q$ has to name a system first.

Why the language goes wrong, in the CED's own words

This confusion is unusual in physics, in that the CED anticipates it by name and devotes a section of its appendix to it. Worth reading, because the exam is written by the same people.

The appendix section is headed Heat vs. Heating vs. Cooling, and it opens by saying that in physics, the term heat has a very specific definition that is a thermodynamic analog to work, and that similar to the amount of work done, heat is the amount of energy transferred into or out of a thermodynamic system through thermal processes such as conduction, convection, or radiation.

It then names the problem directly. It observes that the term can be problematic when used in conversation or in print, because the way heat is used in a sentence does not emphasize that it is an amount of energy being transferred, and that in contexts where you would discuss doing work, the parallel would be heating or cooling.

It is even blunter about what students end up believing. It says that this incorrect use of the word heat is pervasive in many physics classrooms and so students tend to confuse heat with temperature and incorrectly believe objects and systems can have an amount of heat.

And it names its own remedy: because the word heat is so commonly misused and is hard to use naturally in a sentence, heating and cooling are used to help emphasize the processes by which thermal energy is transferred.

That last point explains a feature of the framework worth checking for yourself. Across the whole of Unit 9's required course content, the word heat never appears as a standalone noun meaning transferred energy. Every occurrence is either the gerund, heating or cooling, or the compound term specific heat, which names a material property and not a transfer. 9.3.A.1.i says heating is the transfer of energy into a system by thermal processes, and 9.3.A.1.ii says cooling is the transfer of energy out of a system by thermal processes. 9.4.B.1 describes the first law as accounting for energy transferred into or out of a system by work, heating, or cooling. Verbs throughout. The equation sheet does the same thing in its symbol list, defining $Q$ not as "heat" but as the energy transferred to a system by heating.

So when you write a free-response answer, the CED has already shown you the safe phrasing. Say energy transferred by heating rather than heat, and the sentence cannot make the error.

Heat vs temperature works the other confusion the appendix names, the one that treats heat as a hotness rather than as a stored quantity.

When it costs a mark

• Writing $\Delta Q$. There is no such quantity. $Q$ is already the change in something; it is the amount transferred during a process. A delta in front of it is a sign that the state and the transfer have been merged.
• Answering "how much heat does the gas have" with a number. If a question genuinely asks this, it is asking for the internal energy and using the wrong word, and the safe move is to compute $U = \frac{3}{2}nRT$ and label it internal energy explicitly.
• Assuming $Q > 0$ means the temperature rose. It does not. The straight-line expansion above absorbs $750\ \text{J}$ and cools by nearly $100\ \text{K}$. Only $\Delta U$ tracks temperature, and $\Delta U = Q + W$ has two terms.
• Assuming $\Delta U = Q$ in general. It holds only when $W = 0$, which for a gas means constant volume. 9.4.B.3 calls that case isovolumetric, also known as isochoric. Reaching for it by default converts a two-term equation into a one-term guess.
• Including bulk kinetic energy in $U$. A moving container's internal energy is $\frac{3}{2}nRT$ and nothing else. 9.4.A.2 draws the line: internal structure and behavior on one side, motion of the center of mass on the other.
• Using $U = \frac{3}{2}nRT$ for something that is not an ideal monatomic gas. 9.4.A.1.i is explicit that the form depends on the ideal gas having no internal potential energy. For a solid or a liquid, 9.4.A.1's general statement still applies but that particular formula does not, and the AP Physics 2 sheet gives you $Q = mc\Delta T$ instead.
• Reporting a change in internal energy without a sign. $\Delta U$ is a signed number and its sign is usually the answer. "The internal energy changes by $300\ \text{J}$" is only half of $-300\ \text{J}$.

When they coincide, and why that lulls you

In the one situation students meet most often, the two are numerically equal, and that is exactly why the distinction stays invisible for so long.

Seal a gas in a rigid container so the volume cannot change. Then $\Delta V = 0$, the printed $W = -P\Delta V$ gives $W = 0$, and the first law collapses to

Every joule that arrives by heating shows up as internal energy, one for one. Heat a beaker of water on a hot plate and this is essentially the situation, so the numbers agree in the first experiment anyone runs, and they keep agreeing for as long as nothing moves.

But look closely at what that equation actually says, because even here it does not say the two things are the same. It says the change in one equals the whole of the other. $Q$ is $3739\ \text{J}$ and $\Delta U$ is $3739\ \text{J}$, while $U$ itself is $12{,}465\ \text{J}$, a completely different number that the equation never mentions. The transfer matched the change, not the state. Once the gas has arrived, the $3739\ \text{J}$ of heat is not somewhere inside it; it has stopped existing as a category, and what exists is a larger internal energy.

A second and subtler coincidence sits in the units. Both are joules, both can be written on the same axis, both appear in the same equation. Nothing about the arithmetic warns you, and the arithmetic is right. What differs is what the number is attached to: a moment for $U$, an interval for $Q$.

The fastest way to keep them apart under exam pressure is to ask which question the quantity answers.

• "What is it like now?" is a state question, and internal energy answers it.
• "What happened between then and now?" is a process question, and heat is one of the two answers, work being the other.

A question that gives you a starting state and an ending state and asks for $\Delta U$ needs no information about the path. A question that asks for $Q$ needs the path, and if it has given you one, that is the tell.