Efficiency: Question bank — Physics deep dive

The one rule everything here tests: $η = \frac{E _{useful out}}{E _{total in}} = 1 - \frac{E _{wasted}}{E _{in}}, 0 \leq η \leq 1$

where $η$ (the Greek letter eta, our symbol for efficiency) is a pure fraction: useful energy out divided by total energy in.

True or false — justify

A machine that outputs exactly as much energy as it takes in has

η = 100%

True in principle, but impossible in practice: it would need zero waste (

E_{wasted} = 0

), and every real process leaks some heat, sound or friction. So

η = 100%

is a limit you approach, never reach.

Efficiency can be greater than

100%

for a very well-oiled machine.

False. That would mean

E_{useful out} > E_{in}

, i.e. energy created from nothing — a violation of Conservation of Energy. Oiling reduces friction, pushing

η

up toward

100%

, never past it.

A lever with efficiency near

100%

still "multiplies energy" because a small force lifts a big weight.

False. The lever multiplies force, not energy: the small force moves through a longer distance, so

F \times d

\approx F \times d

out. Energy is not amplified — only traded between force and distance.

If a device wastes no energy at all, its efficiency is

0%

False — it's the opposite. Zero waste means

E_{wasted} = 0

, so

η = 1 - 0 = 100%

. People confuse "no waste" with "no output"; here it means perfect output.

Efficiency has units of joules.

False. It is a ratio of two energies, so the joules cancel —

η

is dimensionless, a bare number between 0 and 1 (or a percentage).

Two machines in series, each

70%

efficient, give an overall efficiency of

140%

False, and impossibly so. Series efficiencies multiply:

0.70 \times 0.70 = 0.49 = 49%

. Adding them would exceed

100%

, which conservation forbids.

A machine can be

100%

efficient at converting electricity into heat.

True — an electric heater. If the heat is the thing you wanted, then all input becomes useful output, so

η \approx 100%

. Efficiency depends entirely on what you call useful.

The wasted

30%

of a motor's energy is destroyed.

False. It is fully conserved, just converted into forms you didn't want — mostly heat in the windings, plus sound and vibration (see Heat and Internal Energy). "Wasted" means "wrong form," not "gone."

Spot the error

"A bulb emits 100 J of light and 400 J of heat, so its efficiency is

500/500 = 100%

The error is counting heat as useful output. For a lamp the useful output is only the light:

η = 100/500 = 20%

. Total output always equals input (conservation) — that tells you nothing about efficiency.

"The pump takes 2 kW and delivers 1400 W, so

η = 2000/1400 = 143%

The ratio is upside-down: efficiency is output÷input, not input÷output. Correct is

1400/2000 = 70%

. Whenever you get

η > 1

, you've flipped the fraction.

"Motor input = 500 kJ, useful = 300 J, so

η = 300/500 = 60%

Units mismatch: 500 kJ = 500000 J. Converting first gives

η = 300/500000 = 0.06%

η

is a ratio, so both quantities must be in the same unit before dividing.

"Efficiency of raising a box is

η = m g h / (electrical input)

, and

m g h

is the input."

The slip is calling

m g h

the input. Here

m g h

is the useful output (gravitational PE gained, see Gravitational Potential Energy); the electrical energy supplied is the input. Output goes on top.

"This engine is

60%

efficient, so it wastes

40%

as heat plus whatever friction takes."

Double-counting. The

40%

waste already includes the friction, sound and heat — everything that isn't useful.

E_{wasted} = E_{in} - E_{useful}

is the total of all losses, not just one of them.

"For chained machines I average the efficiencies:

(80% + 50%) /2 = 65%

Averaging is wrong because the second machine only ever sees the first's reduced output. You multiply:

0.80 \times 0.50 = 40%

— always below the weakest stage, never the average.

Why questions

Why can efficiency never exceed

100%

, no matter how clever the design?

Because

E_{useful out} \leq E_{in}

is forced by Conservation of Energy: output is only part of what came in. Getting

η > 1

would mean creating energy, which no machine can do.

Why do we bother with efficiency if energy is always conserved anyway?

Conservation says the total amount is fixed, but not that it's useful. Efficiency measures how much stays in the form you wanted, so we can compare machines and cut waste — it's bookkeeping about usefulness, not quantity.

Why does friction lower efficiency specifically?

Friction converts ordered kinetic/mechanical energy into scattered heat, which is (usually) not the output you wanted. That heat lands in

E_{wasted}

, so

η = 1 - E_{wasted} / E_{in}

drops.

Why can you use powers instead of energies in the formula?

Because power

P = E / t

, and input and output share the same elapsed time

t

. When you form the ratio the

t

cancels, so

η = P_{useful} / P_{in}

gives the identical number (see Work and Power).

Why is "a good machine runs cool" a fair rule of thumb?

Heat leaving a machine is usually wasted energy. Less heat produced means a smaller

E_{wasted}

, hence a higher

η

. A hot machine is literally radiating away the energy you paid for.

Why do real heat engines have a hard ceiling below

100%

, unlike a motor?

A heat engine must dump some heat to a cold reservoir to keep running — this is set by thermodynamics, not just sloppy engineering. The Carnot limit caps

η

well under 1 even for a perfect frictionless engine (see Heat Engines & Carnot Efficiency).

Edge cases

What is the efficiency of a machine whose useful output is zero (all input becomes heat you didn't want)?

η = 0/ E_{in} = 0%

. This is the lower bound: energy is still conserved, but none of it did the intended job.

Can efficiency be exactly

0%

and exactly

100%

0%

is achievable (a device producing no useful output).

100%

is only a theoretical limit — it requires zero waste, which real friction and heat loss forbid, so real

η

sits strictly between.

If you redefine what counts as "useful," does a machine's efficiency change?

Yes — dramatically. A heater is

\sim 100%

efficient if heat is useful, but near

0%

if you wanted light. Efficiency is defined relative to your goal, so always state what "useful" means first.

A single machine is

90%

efficient. You add a second, perfect,

100%

stage after it. What is the overall efficiency?

0.90 \times 1.00 = 90%

. A perfect stage neither helps nor hurts — multiplying by 1 leaves the total unchanged, confirming the series-multiplication rule.

Ten identical stages, each

95%

efficient, are chained. Is the whole thing still highly efficient?

No:

0.9 5^{10} \approx 0.60 = 60%

. Small losses compound multiplicatively, so many "good" stages in series can add up to a mediocre system — the tyranny of chained efficiency.

What happens to

η

as a machine's wasted energy approaches the full input?

E_{wasted} \to E_{in}

, the formula

η = 1 - E_{wasted} / E_{in} \to 1 - 1 = 0

. The efficiency smoothly falls to

0%

— the limiting case of a completely useless machine.

Recall One-line summary of every trap here

The three deadly mistakes are: (1) flipping the ratio (input÷output), (2) mixing units, and (3) miscounting "useful" — always ask what did I actually want? And two conservation facts pin everything: $η \leq 100%$ always, and series efficiencies multiply.

Connections

Efficiency — parent note; the definition and derivations these traps probe.
Conservation of Energy — the single law behind " $η$ can't beat 100%."
Work and Power — why the power form gives the same number.
Gravitational Potential Energy — the usual "useful output" $m g h$ people misplace.
Heat and Internal Energy — where "wasted" energy actually goes.
Friction — the everyday reason $η < 1$ .
Heat Engines & Carnot Efficiency — why engines have a hard sub-100% ceiling.