The First Law and Where Energy Goes - Physics for Pilot Assessments

This lesson ties the module together with the bookkeeping rule that governs all of it, and then says what that rule does not cover. By the end of it you should be able to define internal energy, use the first law of thermodynamics to find how a gas's internal energy changes when it is heated and does work, explain why heat never flows from cold to hot without something driving it, calculate the efficiency of a heat engine both ways round, and say why no engine can ever turn all of its heat into useful work.

Internal energy

Internal energy is the total energy contained in an object because of its particles — the kinetic energy of all their motion, plus the potential energy stored in the forces holding them near one another. The two lessons on heating and on latent heat each dealt with one of those halves: warming something adds to the kinetic part and shows on a thermometer, while melting or boiling adds to the potential part and does not. Internal energy is written with the letter U, and the crucial difference between it and the other two quantities in this lesson is that it is a property the object has. You could, in principle, state how much internal energy a given cylinder of gas contains at a given moment.

Heat and work are not properties of anything. Both are transfers — energy in the act of crossing from one place to another — so it makes sense to ask how much heat entered a gas during some process, and no sense at all to ask how much heat the gas has now. Once the energy has arrived, it is internal energy, and there is nothing about it to record whether it came in by heating or by having work done on it. That is the point the whole of this lesson rests on.

Two ways in, one bookkeeping rule

So there are exactly two ways to change the internal energy of a gas, and both have already appeared. You can heat it or cool it, which transfers energy because of a temperature difference. Or you can do work on it or let it do work, which transfers energy by moving a piston, as the previous lesson set out. Add up the two contributions with the right signs and you have the change in internal energy — which is nothing more than conservation of energy from the energy module, written out for a gas.

The first law of thermodynamics

\Delta U = Q - W

$\Delta U$: the change in the internal energy of the gas, in joules (J) — positive means it gained
$Q$: the heat added to the gas, in joules (J) — negative if heat left it
$W$: the work done by the gas on its surroundings, in joules (J) — negative if work was done on it instead

The signs are where marks are lost, so read them once slowly. Heat added counts as positive, because it puts energy in. Work done by the gas counts as positive and is then subtracted, because energy the gas spends pushing something else is energy it no longer has. So if the gas has work done on it, W is negative and subtracting a negative adds — which is exactly the compression result from the previous lesson, where an insulated gas with no heat entering it still gained internal energy and warmed up. Be aware that some books define W as the work done on the gas and write the law with a plus sign; the physics is identical, but if you mix the two conventions you will get the sign of your answer wrong, so stick to this one.

Four combinations cover everything. Heat a gas and stop it expanding, and all the heat becomes internal energy, so its temperature rises. Heat a gas and let it expand while doing work, and only the remainder becomes internal energy, so its temperature rises less than you might expect. Compress an insulated gas, and with no heat involved at all the work becomes internal energy and it warms. Let an insulated gas expand and do work, and with no heat coming in the work must come out of its internal energy, so it cools.

Heat only flows one way on its own

The first law is silent on direction, and that is a real gap in it rather than an oversight. Consider a hot cup of coffee cooling in a cool room. Energy leaves the coffee and enters the room, and the total is conserved. Now imagine the reverse: the room gives up a small amount of energy and the coffee spontaneously heats back up. That would conserve energy perfectly — the books balance exactly as well as they do in the real version — and yet it never happens. So there must be a second rule at work, one that is about direction rather than about amounts. In its simplest form it says that heat flows from a hotter body to a colder one and never the other way round of its own accord. That statement is the second law of thermodynamics.

The particle picture makes it feel less arbitrary. Where a hot body meets a cold one, fast particles are colliding with slow ones, and in such a collision the fast one is far more likely to give up energy than to gain it — not certain, just overwhelmingly likely. Any one collision could go either way. But the number of particles involved is of the order of a hundred thousand billion billion, so the average outcome is not a tendency, it is a certainty, and energy flows from the fast population to the slow one until the two averages match. What makes the reverse process impossible in practice is not that any individual step is forbidden but that all of them would have to happen to cooperate at once.

Heat engines, and why some waste is unavoidable

A heat engine is any machine that takes heat from something hot and turns part of it into useful work: a petrol or diesel engine, a steam turbine, a jet engine, a power station. The pattern is always the same three steps. Heat is taken in from a hot source. Some fraction of it comes out as work — a turning shaft, a moving piston, a jet of exhaust. And the rest is dumped into something colder, which in practice means the atmosphere, a river or a cooling tower. That third step is the one people want to design away, and it cannot be done.

The reason is that an engine has to work in a cycle. To produce work over and over again, the gas inside it must be brought back to the state it started in, ready to go round again — and bringing it back means recompressing it, which costs work. If you recompressed it while it was still hot you would spend as much work as you gained, and the engine would achieve nothing. The only way to make the return trip cheap is to cool the gas down first, and cooling it means giving heat to something colder. So the cold sink is not a leak in an otherwise perfect machine; it is the thing that makes the machine possible at all, and the heat sent there is unavailable for work by construction.

Thermal efficiency of a heat engine

\text{efficiency} = \frac{W_\text{useful}}{Q_\text{in}}

$\text{efficiency}$: a bare ratio between 0 and 1, with no units — multiply by 100 for a percentage
$W_\text{useful}$: the useful work the engine delivers, in joules (J)
$Q_\text{in}$: the heat supplied to it from the hot source, in joules (J)

Since the energy supplied has only two places to go, the heat rejected is simply whatever the useful work was not — which gives you a free check on any answer, and a second route to the same figure.

The widths carry the argument. Efficiency is the narrow branch divided by the flow in, and most of the energy supplied to a real engine leaves down the wide branch instead — some heat must always be rejected, and no amount of engineering closes that branch off entirely.

Real figures put the point beyond argument. A petrol engine converts around 25 to 30 per cent of the energy in its fuel into work at the crankshaft, a large diesel 35 to 45 per cent, a coal-fired power station about 40, and the best combined-cycle gas stations around 60 — that last one being unusually good precisely because it catches the waste heat of one stage and uses it to drive a second. Everything not in those percentages goes into the air, the cooling water or the exhaust as heat. That is where the phrase waste heat comes from, and it is worth being clear that nothing is destroyed: the energy is all still there, it is simply spread thinly through cool surroundings where nothing can be done with it.

There is a ceiling on how well any engine can possibly do, and it depends only on the two temperatures it works between. The hotter the source and the colder the sink, the higher the ceiling — which is why engines are run as hot as their materials will stand.

The best efficiency any heat engine could have

\text{efficiency}_\text{max} = 1 - \frac{T_\text{cold}}{T_\text{hot}}

$\text{efficiency}_\text{max}$: the theoretical ceiling, as a bare ratio between 0 and 1
$T_\text{cold}$: the absolute temperature of the cold sink, in kelvin (K)
$T_\text{hot}$: the absolute temperature of the hot source, in kelvin (K)

Notice what that expression rules out. The fraction can only vanish, giving a ceiling of 1, if the cold sink is at absolute zero — which is unreachable — so the ceiling is always below 100 per cent, and real engines fall well short of even that. No real machine reaches this limit, because it assumes there is no friction and no unwanted heat leakage anywhere in the design.

Refrigeration: pumping heat the wrong way

Heat will not flow from cold to hot on its own, and the words on its own are load-bearing: it will do so perfectly well if something drives it. That is what a refrigerator is for, and it uses nothing but the compression and expansion effects from the previous lesson. A compressor squeezes the working gas, which warms it — well above room temperature. That hot gas is passed through the fins on the back or underneath of the cabinet, where it is hotter than the kitchen and so gives up heat to it in the ordinary direction. Then the gas is allowed to expand, which cools it well below the temperature inside the cabinet, and it is passed through pipes in there, where it is now the colder body and so absorbs heat from the food in the ordinary direction. Round it goes again.

So no law has been broken at any stage: every actual transfer runs from hotter to colder, and the trick is that the gas is made hotter than the room in one place and colder than the food in another. What it costs is the work the compressor does, which is what the electricity supply pays for — and that is why the second law is often stated as saying that heat can be moved from cold to hot only at the cost of work. Two consequences follow, and both are examined. First, all of the energy involved ends up in the kitchen: the heat taken out of the food, plus the electrical energy the compressor used, comes out of those fins, so a refrigerator is a net heater of the room it stands in and leaving its door open makes the kitchen warmer, not cooler. Second, the same machine run for the sake of its warm end rather than its cold one is a heat pump, and because it is moving heat rather than generating it, it delivers more heat into a building than the electrical energy it consumes.

Example 1 — efficiency from the energy flows

An engine is supplied with 1000 joules of heat from its hot source and delivers 300 joules of useful work. What is its efficiency, and how much energy is rejected as heat?

Efficiency is useful work divided by energy supplied: $\text{efficiency} = \frac{300}{1000} = 0.3$
As a percentage that is 0.3 × 100 = 30 per cent.
The energy supplied has only two destinations, work and rejected heat, so the rejected heat is what is left: $1000 - 300 = 700\ \text{J}$
Check the shares add up: 300 joules of work plus 700 joules of waste is 1000 joules in, which is what was supplied. Nothing has been created or lost.
So a 30 per cent efficient engine rejects 70 per cent of everything it is given, which is roughly what a petrol engine actually does.

Example 2 — running the efficiency backwards

An engine is 25 per cent efficient and has to deliver 500 joules of useful work. How much heat must be supplied to it, and how much will it reject?

Write the efficiency as a decimal first: 25 per cent is $\frac{25}{100} = 0.25$ .
Start from $\text{efficiency} = \frac{W_\text{useful}}{Q_\text{in}}$ and multiply both sides by the energy supplied, then divide both sides by the efficiency. That leaves $Q_\text{in} = \frac{W_\text{useful}}{\text{efficiency}}$ .
So the heat supplied is $\frac{500}{0.25}$ . Dividing by 0.25 is the same as multiplying by 4, so it is 2000 joules.
The rejected heat is the remainder: 2000 − 500 = 1500 joules.
Sense-check the direction. A less efficient engine needs more fuel for the same work, and 25 per cent efficiency needing four times the energy it delivers is exactly that.

Example 3 — the first law, both ways round

First: 500 joules of heat are supplied to a gas, and the gas expands and does 200 joules of work on a piston. What is the change in its internal energy? Second: 300 joules of work are done on a gas while 100 joules of heat leak out of it. What is the change in its internal energy now?

For the first case, identify the signs. Heat was added, so Q = +500 joules. The gas did the work, so W = +200 joules.
Apply change in internal energy = heat in − work done by the gas: $\Delta U = 500 - 200 = +300\ \text{J}$ The gas gained 300 joules, so it ends up warmer than it started.
For the second case, both signs flip. Heat left the gas, so Q = −100 joules. Work was done on the gas rather than by it, so W = −300 joules.
Apply the law again: $\Delta U = (-100) - (-300)$ Subtracting a negative is adding, so this is −100 + 300 = +200 joules.
So the gas gained internal energy and warmed up, even though heat was flowing out of it the whole time — because the work being done on it more than paid for the leak.
That second result is the one worth remembering. Losing heat does not guarantee cooling, if work is being done on the gas at the same time.

Example 4 — the ceiling nobody reaches

An engine takes heat from a source at 600 K and rejects it to surroundings at 300 K. What is the best efficiency it could possibly have? If it actually manages 30 per cent, how does that compare?

Both temperatures are already in kelvin, which the relationship requires — a Celsius figure here would be meaningless, exactly as in the gas laws.
The ceiling is $1 - \frac{T_\text{cold}}{T_\text{hot}} = 1 - \frac{300}{600}$ .
$\frac{300}{600} = 0.5$ , so the ceiling is 1 − 0.5 = 0.5, which is 50 per cent.
The real engine manages 30 per cent, so it is achieving $\frac{30}{50} = 0.6$ , or 60 per cent of the best that was ever available to it. The other 40 per cent of the possible is lost to friction and unwanted heat leakage.
Note that even the perfect version rejects half of everything it is given. Raising the ceiling means either a hotter source or a colder sink, and the sink is usually the outside air, which you do not get to choose.

Why it matters in aviation

An aero engine is a heat engine, and this bookkeeping is why it burns as much fuel as it does rather than something better. Only part of the energy released by the fuel emerges as useful work; the rest leaves in the exhaust stream and in the air and oil used for cooling, and it leaves because the cycle cannot be completed without rejecting it. That is what makes thermal efficiency a figure worth quoting for an engine at all, instead of an admission of poor workmanship.

It also explains a result that looks odd at first. The engine works between the temperature of its hottest gas and the temperature of the air it rejects heat to, and the air at cruising level is very cold, so the ceiling on what the same engine could achieve is higher up there than it is standing on a warm apron. And it explains where the design effort goes. Since that ceiling rises with the temperature of the hot end, engines are run as hot as their materials will bear, and the temperature at the hottest point in the cycle is watched closely, because it is at once the source of the efficiency and the limit on how long the engine lasts.

What the test asks

Two distractors, both attacking the second law. The first is the perfect engine: an option in which all of the heat supplied is converted into useful work, or an efficiency of 100 per cent, sometimes justified by saying that energy is conserved so none of it need be wasted. Energy being conserved is exactly why the first law permits it and the second law forbids it — an engine must reject heat to a colder place in order to complete its cycle, so its efficiency is always well below 100 per cent, and any option offering the full amount is wrong on sight. The second is heat flowing from cold to hot with nothing driving it, usually phrased as a cold object giving up energy to a warmer one, or as a room warming a cup of coffee. It happens only in a refrigerator, and only because a compressor is doing work to make it happen. Expect too a sign error in a first-law item, where the answer offered is the one you get by adding the work instead of subtracting it — check whether the work was done by the gas or on it before writing anything down. And on the refrigerator, the reliable trap is the claim that an open fridge door cools a room. It warms it, because the compressor's work ends up in the room as well.