Q.A Carnot engine whose efficiency is 45% takes heat from a source maintained at a temperature of 327°C. To have an engine of efficiency 60% for the same exhaust (sink) temperature, what must the intake temperature be?
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Carnot Efficiency: The Ultimate Limit
Imagine you have a heat engine — a device that takes in heat from a hot source, does some useful work (like turning a wheel or generating electricity), and dumps the leftover heat into a cold sink. A steam engine, a car engine, a power plant — they all do this.
Your first question should be: How much of the heat I put in can I actually turn into work? That fraction is the efficiency. If you put in 100 J of heat and get 40 J of work out, the efficiency is 0.4 (or 40%). The rest — 60 J — is wasted heat.
Now, can you ever get 100% efficiency? Turn all the heat into work, with zero waste? Intuition says no. Heat flows spontaneously from hot to cold, not the other way. To extract work, you must let some heat fall to the cold sink — that's the price of doing business. The question is: what's the best you can possibly do?
That's where Carnot comes in.
The Carnot Engine: A Perfect, Idealized Machine
In 1824, Sadi Carnot imagined a perfectly reversible engine — one that operates without any friction, heat loss, or wasteful turbulence. It's a theoretical ideal, not something you can build. But it sets the absolute upper limit on efficiency for any heat engine working between two temperatures.
The Carnot engine works in a cycle of four reversible steps: two isothermal (constant temperature) and two adiabatic (no heat exchange). The details of the cycle matter less than the result.
ηCarnot=1−ThTc
Here:
- Th = absolute temperature of the hot reservoir (source), in Kelvin
- Tc = absolute temperature of the cold reservoir (sink), in Kelvin
That's it. The efficiency depends only on the two temperatures. Nothing else — not the working substance, not the design, not the size.
What This Tells You
First, notice the fraction Tc/Th. If the cold sink is at absolute zero (0 K), then Tc/Th=0 and efficiency = 1 (100%). But you can never reach absolute zero — that's the third law of thermodynamics. So 100% efficiency is impossible.
Second, the bigger the temperature difference, the higher the efficiency. A hot source at 600 K and a cold sink at 300 K gives η=1−300/600=0.5 (50%). If you raise the hot source to 900 K (same cold sink), η=1−300/900≈0.667 (66.7%). Hotter source = better efficiency.
Third, no real engine can beat this. A steam turbine, a car engine, a jet engine — all have efficiencies lower than the Carnot limit for their operating temperatures. The Carnot efficiency is the ceiling.
A common mistake: thinking Carnot efficiency depends on the amount of heat or the type of fuel. It does not. Only the two temperatures matter. A coal plant and a nuclear plant operating between the same Th and Tc have the same Carnot limit — even though one burns coal and the other splits atoms.
Why Only Temperatures? …
From η=45% at TH=600 K, the fixed sink temperature is TL=330 K; requiring η=60% at the same sink gives a new source t …
Step 1. Convert the source temperature to kelvin: TH=327°C+273=600 K. Using the given efficiency η=0.45=1−TL/TH, solve for the (fixed) sink temperature: TL=TH(1−η)=600×(1−0.45)=600×0.55=330 K. …
First back out the fixed sink temperature from the given 45% efficiency, then use it with the new 60% ta …
- Forgetting the sink temperature is meant to stay THE SAME between the two scenarios ('for the same exhaust temperature') -- both efficien …
- CBSE 2026Set ANNUAL1 markMCQQ.If the hot reservoir temperature of a Carnot engine is 800 K and the efficiency is 25%, then the temperature of cold reservoir will be(a) − 173.15°C(b) − 73.15°C(c) 26.85°C(d) 126.85°C
›Reveal solutionSolution
Carnot: Tc = Th(1 - eta) = 800 x 0.75 = 600 K = 326.85 degrees C; no listed option matches this.
The efficiency of a Carnot engine is eta = 1 - Tc/Th, where Th and Tc are the hot- and cold-reservoir temperatures in kelvin.
Given Th = 800 K and eta = 25% = 0.25:
Tc/Th = 1 - eta = 1 - 0.25 = 0.75
Tc = 0.75 x 800 = 600 K.
Converting to Celsius: Tc = 600 - 273.15 = 326.85 degrees C.
…
- CBSE 2025Set ANNUAL1 markMCQQ.An ideal engine works between 327°C and 27°C. What is its efficiency? (A) 60% (B) 80% (C) 40% (D) 50%
›Reveal solutionSolution
The ideal engine's efficiency working between 600 K and 300 K is 50%.
Converting to Kelvin: T1=327+273=600 K (source), T2=27+273=300 K (sink).
Carnot (ideal) engine efficiency:
…
- CBSE 2024Set ANNUAL1 markMCQQ.Even a Carnot engine cannot give 100% efficiency because we cannot(a) prevent radiation(b) find ideal sources(c) reach absolute zero temperature.(d) eliminate friction
›Reveal solutionSolution
Carnot efficiency η=1−T2/T1 becomes 100% only when the sink is at T2=0K, which is unattainable in practice.
For a Carnot engine operating between a source at temperature T1 and a sink at temperature T2, the efficiency is η=1−T1T2. This equals 1 (100%) only when T2=0K, i.e. the sink is at absolute zero. By the third law of thermodynamics, absolute zero can never actually be reached, so no real (or even ideal Carnot) engine can attai …
- CBSE 2022Set ANNUAL1 markMCQQ.The efficiency of a heat engine working between the freezing point and boiling point of water is :(a) 26.8%(b) 6.25%(c) 12.5%(d) 20%
›Reveal solutionSolution
The efficiency of an ideal heat engine operating between two fixed temperatures is the Carnot efficiency, η = 1 − (Tc/Th), using absolute (Kelvin) temperatures for the cold and hot reservoirs.
Convert the given temperatures to Kelvin:
Freezing point of water (cold reservoir): Tc = 0°C = 273 K
Boiling point of water (hot reservoir): Th = 100°C = 373 K
Carnot efficiency formula:
η = 1 − (Tc / Th)
Substitute values:
η = 1 − (273/373) = (373 − 273)/373 = 100/373
η ≈ 0.2681 = 26.8%
…
- CBSE 2022Set ANNUAL1 markQ.Write a formula for the efficiency of a Carnot engine.
›Reveal solutionSolution
The efficiency of a Carnot engine is η = 1 - T2/T1, where T1 is the source (hot reservoir) temperature and T2 is the sink (cold reservoir) temperature, both in kelvin.
For a Carnot cycle operating between a source at temperature T1 and a sink at temperature T2, absorbing heat Q1 from the source and rejecting heat Q2 to the sink, the efficiency is defined as the ratio of net work output to heat input: η = W/Q1 = (Q1-Q2)/Q1. For a reversible Carnot cycle it can be shown tha …
- CBSE 2022Set ANNUAL1 markMCQQ.An ideal engine work between two temperature 327°C (source) and 27°C (sink). What is the efficiency ?(a) 60%(b) 80%(c) 40%(d) 50%
›Reveal solutionSolution
The Carnot (ideal) efficiency is 50%.
Convert temperatures to kelvin:
- Source T₁ = 327 + 273 = 600 K
- Sink T₂ = 27 + 273 = 300 K …
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