Why must the temperatures in the Carnot efficiency formula be in Kelvin?

The formula η = 1 − T₂/T₁ uses a ratio of absolute (thermodynamic) temperatures. The Kelvin scale is the only common scale on which the ratio Q₂/Q₁ equals T₂/T₁, because the Carnot relation Q₁/Q₂ = T₁/T₂ itself defines the absolute temperature scale. Plugging in Celsius gives a wrong ratio and a wrong efficiency. Always convert: T(K) = T(°C) + 273.

Does the Carnot efficiency depend on the working substance?

No. Carnot's theorem proves the efficiency of a Carnot engine is independent of the nature of the working substance. NCERT derives η = 1 − T₂/T₁ using an ideal gas only because it has a simple equation of state, but the result holds for any reversible engine operating between the same two reservoirs. Efficiency depends only on T₁ and T₂.

What are the four steps of the Carnot cycle, in order?

Step 1→2: isothermal expansion at T₁, absorbing heat Q₁ from the hot reservoir. Step 2→3: adiabatic expansion, temperature falls from T₁ to T₂. Step 3→4: isothermal compression at T₂, releasing heat Q₂ to the cold reservoir. Step 4→1: adiabatic compression, temperature rises from T₂ back to T₁. Two isothermals connected by two adiabatics, all reversible.

Can a Carnot engine ever reach 100% efficiency?

Only in principle, and only if T₂ = 0 K. From η = 1 − T₂/T₁, efficiency equals 1 (i.e. 100%) when the sink temperature T₂ is absolute zero. Absolute zero is unattainable, so no real or ideal engine reaches 100%. This is the practical content of the second law: η can never equal one.

How do I raise the efficiency of a Carnot engine?

Either raise the source temperature T₁ or lower the sink temperature T₂, because η = 1 − T₂/T₁ grows as the ratio T₂/T₁ shrinks. For the same temperature difference, lowering T₂ raises efficiency more than raising T₁ by the equal amount, since T₂ also sits in the denominator-adjacent role. The widest possible T₁−T₂ gap gives the highest efficiency.

What is the coefficient of performance of a Carnot refrigerator?

Running a Carnot engine in reverse gives a Carnot refrigerator. Its coefficient of performance is COP = Q₂/W = T₂/(T₁ − T₂), where T₂ is the cold space and T₁ the hot surroundings. The COP can be greater than one and rises sharply as T₁ and T₂ approach each other; it becomes infinite only when T₁ = T₂, which the second law forbids as a useful refrigerator.

Why must the two non-isothermal steps of the Carnot cycle be adiabatic?

The engine works between only two reservoirs, at T₁ and T₂. To change the working substance's temperature reversibly between T₁ and T₂ without a third reservoir, no heat may be exchanged during those steps — that is exactly an adiabatic process. Any non-adiabatic step (e.g. isochoric) would need a continuous series of intermediate reservoirs to stay quasi-static, contradicting a two-reservoir engine.

Carnot Engine & Efficiency — NEET Physics

Why a reversible engine sets the limit

A heat engine draws heat $Q_1$ from a hot reservoir, converts part of it to work $W$, and dumps the remainder $Q_2$ into a cold reservoir. The second law of thermodynamics forbids $Q_2 = 0$: no engine can convert heat entirely into work. The question Carnot answered is the next one — given that some heat must be wasted, what is the least we can possibly waste, and which cycle achieves it?

NCERT's reasoning is that the ideal engine must be a reversible engine. Irreversibility springs from dissipative effects — friction, viscosity, finite-temperature heat flow — and every dissipative step throws away usable work, lowering efficiency. A process is reversible only if it is quasi-static and non-dissipative. Because a finite temperature difference between system and reservoir is itself a source of irreversibility, heat must be absorbed from the hot reservoir isothermally at $T_1$ and released to the cold reservoir isothermally at $T_2$.

That fixes two of the four steps. To carry the working substance between $T_1$ and $T_2$ — and back — without any third reservoir, the only reversible option is a step that exchanges no heat at all: an adiabatic process. Any other choice (say an isochoric step) would demand a continuous ladder of intermediate reservoirs to stay quasi-static, which contradicts an engine running between only two temperatures. Two isothermals joined by two adiabatics: that is the Carnot cycle, and a reversible engine running it is a Carnot engine.

The four steps of the Carnot cycle

NCERT takes an ideal gas as the working substance and labels the cycle's corners $1 \to 2 \to 3 \to 4 \to 1$. Each step is reversible, and the four together return the gas to its starting state so that $\Delta U = 0$ over the cycle. The parallel structure of the four steps is best read as a flow.

Step	Process	State change	Heat & work
1 → 2	Isothermal expansion at $T_1$	$(P_1,V_1,T_1) \to (P_2,V_2,T_1)$	Absorbs $Q_1$ from hot reservoir; $W_{1\to2}=Q_1=\mu R T_1 \ln\!\dfrac{V_2}{V_1}$
2 → 3	Adiabatic expansion	$(P_2,V_2,T_1) \to (P_3,V_3,T_2)$	No heat exchange; $W_{2\to3}=\dfrac{\mu R (T_1-T_2)}{\gamma-1}$, temperature falls $T_1 \to T_2$
3 → 4	Isothermal compression at $T_2$	$(P_3,V_3,T_2) \to (P_4,V_4,T_2)$	Releases $Q_2$ to cold reservoir; $W_{3\to4}=Q_2=\mu R T_2 \ln\!\dfrac{V_3}{V_4}$
4 → 1	Adiabatic compression	$(P_4,V_4,T_2) \to (P_1,V_1,T_1)$	No heat exchange; $W_{4\to1}=\dfrac{\mu R (T_1-T_2)}{\gamma-1}$, temperature rises $T_2 \to T_1$

The two adiabatic work terms are equal in magnitude and opposite in role — the work the gas does expanding from $T_1$ to $T_2$ is exactly returned to the gas when it is compressed from $T_2$ back to $T_1$. They cancel in the cycle total, which is why the net work depends only on the two isothermal steps. That cancellation is the engine of the whole derivation.

The P-V diagram of the cycle

Plotted on a pressure–volume diagram, the cycle is a closed loop bounded by two isotherms (the upper at $T_1$, the lower at $T_2$) and two steeper adiabatics. The loop is traversed clockwise, so the enclosed area is the net work output $W$ per cycle.

Figure 1. The Carnot cycle on a P-V diagram. Red isotherm $A\to B$ at $T_1$ absorbs $Q_1$; purple adiabatics drop and raise the temperature; teal isotherm $C\to D$ at $T_2$ rejects $Q_2$. The clockwise loop encloses the net work $W$. (NCERT Fig. 11.9, with corners relabelled $A,B,C,D$ for $1,2,3,4$.)

The adiabatics are steeper than the isotherms at every point because their slope carries the extra factor $\gamma$: for an adiabatic $PV^{\gamma}=$ const, whereas for an isotherm $PV=$ const. That geometric fact is itself a recurring NEET question, and it is why the two adiabatic arcs cut across the isotherms rather than running parallel to them.

Related drill

The isothermal and adiabatic curve equations come straight from thermodynamic processes — revise $PV=$ const vs $PV^{\gamma}=$ const before tackling cycle areas.

Deriving η = 1 − T₂/T₁

The efficiency of any heat engine is the work output divided by the heat input. For the Carnot cycle the adiabatic work cancels, so the net work is the difference of the two isothermal contributions, and the heat input is $Q_1$. NCERT writes

$$\eta = \frac{W}{Q_1} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2 \ln(V_3/V_4)}{T_1 \ln(V_2/V_1)}.$$

The logarithmic terms are not yet equal, so this is not the final answer. They are tied together by the two adiabatic steps. Applying $TV^{\gamma-1}=$ const to step $2\to3$ and to step $4\to1$:

$$\frac{V_2}{V_1} = \frac{V_3}{V_4}.$$

This is NCERT Eq. (11.26): the two volume ratios are identical, so the logarithms cancel. The efficiency collapses to a statement about temperatures alone:

$$\boxed{\;\eta = 1 - \dfrac{T_2}{T_1}\;}\qquad\text{(Carnot engine, } T \text{ in Kelvin)}$$

Three readings of this single equation carry almost every NEET question on the topic. First, the efficiency depends only on the two reservoir temperatures — not on the working substance, the pressure, or the volumes. Second, because $T_2 < T_1$ always, $\eta$ is always less than one. Third, the cancellation of the volume ratios also yields the universal relation $\dfrac{Q_1}{Q_2} = \dfrac{T_1}{T_2}$, valid for any Carnot engine and used to define the absolute thermodynamic temperature scale.

Worked example

A Carnot engine operates between the boiling point and freezing point of water. Find its efficiency.

Convert to Kelvin. $T_1 = 100^\circ\text{C} = 373~\text{K}$, $T_2 = 0^\circ\text{C} = 273~\text{K}$.

Apply the formula. $\eta = 1 - \dfrac{273}{373} = 1 - 0.732 = 0.268$, i.e. $\eta \approx 26.8\%$. Even between two such extreme everyday reservoirs, barely a quarter of the absorbed heat becomes work — the rest is dumped into the cold reservoir.

The energy-flow picture

It helps to see the cycle as a one-way cascade of energy. Heat $Q_1$ flows in from the hot reservoir; work $W = Q_1 - Q_2$ is extracted; the leftover $Q_2$ is rejected to the cold reservoir. The fraction of $Q_1$ that escapes as work is the efficiency.

Figure 2. Energy flow through a Carnot engine. Of the heat $Q_1$ drawn from the hot reservoir, the fraction $\eta = 1 - T_2/T_1$ leaves as work $W$; the remainder $Q_2$ is rejected to the cold reservoir. The efficiency is fixed by $T_1$ and $T_2$ alone.

Carnot's theorem

NCERT states the result in two parts, together called Carnot's theorem: (a) working between two given temperatures $T_1$ and $T_2$, no engine can be more efficient than a Carnot (reversible) engine; and (b) the efficiency of the Carnot engine is independent of the working substance. A direct corollary is that all reversible engines operating between the same two reservoirs have exactly the same efficiency.

The proof is a reductio against the second law. Couple a reversible Carnot engine $R$, run backwards as a refrigerator, to a hypothetical irreversible engine $I$ that claims a higher efficiency $\eta_I > \eta_R$. Both work between the same source and sink. Let $I$ absorb $Q_1$ from the source, deliver work $W'$, and reject $Q_1 - W'$ to the sink. Arrange $R$ to return the same $Q_1$ to the source while drawing $Q_2$ from the sink and requiring work $W = Q_1 - Q_2$. If $\eta_I > \eta_R$, then $W' > W$, and the coupled $I$-$R$ system would extract net heat $W' - W$ from the cold sink and convert it entirely to work, with no other change anywhere. That is precisely what the Kelvin–Planck statement forbids.

The contradiction kills the assumption: no engine can beat the Carnot engine. The same coupling argument, run with a reversible engine of a different working substance in place of $I$, shows that no substance can do better than another, so the Carnot efficiency is universal. This is exactly why NCERT is entitled to derive $\eta = 1 - T_2/T_1$ using an ideal gas — the simple equation of state is a calculational convenience, but the result holds for every Carnot engine regardless of substance.

The Carnot refrigerator and COP

Every step of the Carnot cycle is reversible, so the whole cycle can be run backwards. Reversed, the engine becomes a Carnot refrigerator: work $W$ is supplied from outside, heat $Q_2$ is extracted from the cold reservoir at $T_2$, and heat $Q_1 = Q_2 + W$ is delivered to the hot reservoir at $T_1$. The figure of merit is not efficiency but the coefficient of performance, the ratio of the heat removed from the cold space to the work paid in.

$$\text{COP} = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2}.$$

Using the Carnot relation $Q_1/Q_2 = T_1/T_2$, the heats give way to temperatures, exactly as in the engine derivation:

$$\boxed{\;\text{COP} = \dfrac{T_2}{T_1 - T_2}\;}\qquad\text{(Carnot refrigerator)}$$

Unlike efficiency, the COP can exceed one — a good refrigerator pumps several joules of heat for each joule of work. The COP grows without bound as $T_1$ and $T_2$ approach each other, and would become infinite at $T_1 = T_2$; but a refrigerator with no temperature gap does no useful cooling, and the second law (Clausius statement) forbids heat moving from cold to hot with no work at all. The COP and the engine efficiency are linked: $\text{COP} = (1-\eta)/\eta$ for the same two reservoirs.

Why 100% efficiency needs T₂ = 0 K

Set $\eta = 1$ in $\eta = 1 - T_2/T_1$. The only solution is $T_2 = 0~\text{K}$: the sink must be at absolute zero. Equivalently, $Q_2 = 0$ demands $T_2 = 0$, so all the absorbed heat would convert to work. Since absolute zero is unattainable, the Carnot efficiency is always strictly less than one — no engine, real or ideal, ever reaches 100%. This is the quantitative face of the Kelvin–Planck statement that a perfect heat engine is impossible.

Two everyday consequences follow. A practical engine, riddled with friction and finite-temperature heat flow, falls well below the Carnot ceiling for its operating temperatures. And the route to higher real efficiency is metallurgical and thermodynamic — raise the source temperature $T_1$ as far as materials allow, and reject heat to as cold a sink $T_2$ as the environment provides.

Quick recap

Carnot in one breath

Carnot cycle = two reversible isothermals ($1\to2$ at $T_1$, $3\to4$ at $T_2$) joined by two reversible adiabatics ($2\to3$, $4\to1$).
$Q_1$ absorbed isothermally at $T_1$; $Q_2$ rejected isothermally at $T_2$; adiabatic work terms cancel over the cycle.
Volume ratios match ($V_2/V_1 = V_3/V_4$), giving $\eta = 1 - T_2/T_1$ with $T$ in Kelvin.
Efficiency depends only on $T_1$ and $T_2$ — never on the working substance, mass or volume.
Carnot's theorem: no engine beats a reversible engine between the same reservoirs; all reversible engines share the same efficiency.
Run in reverse → Carnot refrigerator with $\text{COP} = T_2/(T_1 - T_2)$; also $Q_1/Q_2 = T_1/T_2$.
$\eta = 100\%$ only if $T_2 = 0~\text{K}$, which is unattainable — so $\eta < 1$ always.

Carnot Engine & Efficiency