{"id":721,"date":"2026-08-06T15:56:14","date_gmt":"2026-08-06T15:56:14","guid":{"rendered":"https:\/\/physicsfundamentalsinfo.com\/blog\/?p=721"},"modified":"2026-08-24T13:03:50","modified_gmt":"2026-08-24T13:03:50","slug":"carnot-efficiency","status":"publish","type":"post","link":"https:\/\/physicsfundamentalsinfo.com\/blog\/thermodynamics\/carnot-efficiency\/","title":{"rendered":"Carnot Efficiency: The Efficiency Ceiling for Heat Engines"},"content":{"rendered":"\n
Definition<\/div>

\n \nCarnot efficiency is the maximum fraction of heat energy that any engine can convert into work while operating between two fixed temperatures. It equals one minus the cold reservoir temperature divided by the hot reservoir temperature, with both measured in kelvin. No real engine, however well built, can exceed this limit.\n \n<\/p><\/div>\n \n

Drive past a power station and you will see white plumes drifting from the cooling towers. That is not pollution. That is the plant deliberately throwing away roughly half the energy it just paid for, because physics gives it no other option.<\/p>\n \n

It is not bad engineering either. It is a ceiling worked out in 1824 by a French engineer in his twenties, before anyone knew what heat actually was \u2014 and two centuries of materials science have not moved it by a single percentage point.<\/p>\n \n

\n \n \"Cooling\n \n
Cooling towers are the cold reservoir made visible. Every heat engine needs one.<\/figcaption>\n \n<\/figure>\n \n

What Is Carnot Efficiency?<\/h2>\n \n

Carnot efficiency is the highest thermal efficiency any engine can reach when it runs between a hot reservoir at temperature Th<\/sub> and a cold reservoir at temperature Tc<\/sub>. It is a ceiling, not a target \u2014 and it depends on nothing except those two temperatures.<\/p>\n \n

Strip any heat engine down and it does the same three things. It takes heat in from something hot, turns part of that heat into useful work<\/a>, and dumps the leftover into something cold.<\/p>\n \n

That third step is the one students want to delete. You cannot. Without somewhere cold to reject heat into, the working fluid never returns to its starting state and the engine cannot run a second cycle.<\/p>\n \n

So the honest question is never “how do I stop wasting heat?” It is “what is the smallest fraction I am forced to throw away?” Carnot answered exactly that.<\/p>\n \n

Carnot’s Theorem, in Plain Words<\/h3>\n \n

Sadi Carnot proved two claims that still stand. First: no engine working between two given reservoirs can beat a reversible engine working between the same two. Second: every reversible engine between those reservoirs has the same<\/em> efficiency, regardless of how it is built or what fluid is inside it.<\/p>\n \n

Together those give a universal speed limit. It applies to a steam turbine, a jet engine, a Stirling engine and a hypothetical machine nobody has invented yet.<\/p>\n \n

\"Carnot<\/figure>\n \n

Every heat engine sits between two reservoirs. Efficiency is W divided by Qh<\/sub> \u2014 and Qc<\/sub> can never be zero.<\/p>\n \n

The Carnot Efficiency Formula<\/h2>\n \n

The Carnot efficiency formula is one minus the ratio of the two absolute temperatures. Here it is:<\/p>\n \n

\u03b7 = 1 \u2212 Tc \/ Th<\/div>\n \n

Two rearrangements are worth memorising, because exam questions and real design work use them constantly:<\/p>\n \n

\u03b7 = (Th \u2212 Tc) \/ Th<\/div>\n \n
Tc = Th \u00d7 (1 \u2212 \u03b7)<\/div>\n \n

Every Symbol, With Its SI Unit<\/h3>\n \n
\n\n\n\n\n\n\n\n\n\n
Symbol<\/th>\nQuantity<\/th>\nSI unit<\/th>\n<\/tr>\n<\/thead>\n
\u03b7<\/strong> (eta)<\/td>\nCarnot efficiency \u2014 the maximum work-per-unit-heat<\/td>\ndimensionless, 0 to 1 (multiply by 100 for %)<\/td>\n<\/tr>\n
Th<\/sub><\/strong><\/td>\nAbsolute temperature of the hot reservoir<\/td>\nkelvin (K)<\/td>\n<\/tr>\n
Tc<\/sub><\/strong><\/td>\nAbsolute temperature of the cold reservoir<\/td>\nkelvin (K)<\/td>\n<\/tr>\n
Qh<\/sub>, Qc<\/sub><\/strong><\/td>\nHeat absorbed from hot, heat rejected to cold, per cycle<\/td>\njoule (J)<\/td>\n<\/tr>\n
W<\/strong><\/td>\nNet work output per cycle, W = Qh<\/sub> \u2212 Qc<\/sub><\/td>\njoule (J)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n \n

How Do You Calculate Carnot Efficiency?<\/h3>\n \n

Three steps, in this order:<\/p>\n \n

    \n
  1. Convert both temperatures to kelvin.<\/strong> T(K) = T(\u00b0C) + 273.15. Skip this and your answer will be wrong, often by more than double.<\/li>\n
  2. Divide the cold temperature by the hot one.<\/strong> That ratio Tc<\/sub>\/Th<\/sub> is the fraction you are forced to throw away.<\/li>\n
  3. Subtract from 1.<\/strong> Multiply by 100 if you want a percentage.<\/li>\n<\/ol>\n \n

    Worth doing by hand a few times so the kelvin step becomes automatic. After that, our Carnot Efficiency Calculator<\/a> will do the conversion and the arithmetic for you, and it also rearranges the formula to solve for either reservoir temperature when you already know the efficiency you are aiming at.<\/p>\n \n

    Why Kelvin, and Only Kelvin<\/h3>\n \n

    The formula divides one temperature by another, so the zero point of your scale is not cosmetic \u2014 it changes the answer completely. Celsius puts zero at the freezing point of water, which is an arbitrary place to start counting.<\/p>\n \n

    Kelvin starts at absolute zero, the genuine bottom of the temperature scale, which is why NIST describes it as an absolute scale<\/a> where 0 K equals \u2212273.15 \u00b0C. Only on that scale does the ratio Tc<\/sub>\/Th<\/sub> mean anything physical.<\/p>\n \n

    A quick sanity check that catches most slips: if your engine’s efficiency comes out above about 90%, you almost certainly forgot to convert. Real reservoir pairs rarely allow it. If you are hazy on why absolute temperature behaves so differently from everyday heat, the difference between heat and temperature<\/a> is the thing to fix first.<\/p>\n \n

    How the Carnot Cycle Reaches That Limit<\/h2>\n \n

    The Carnot cycle hits the ceiling by using only four reversible steps \u2014 two isothermal, where heat crosses at a vanishingly small temperature difference, and two adiabatic, where no heat crosses at all. Nothing is ever wasted by pushing heat down a temperature gap.<\/p>\n \n

    Follow a gas through one full loop:<\/p>\n \n

      \n
    1. a to b \u2014 isothermal expansion at Th<\/sub>.<\/strong> The gas touches the hot reservoir, absorbs Qh<\/sub> and expands while staying at Th<\/sub>, pushing the piston out.<\/li>\n
    2. b to c \u2014 adiabatic expansion.<\/strong> The reservoir is removed. The gas keeps expanding and cools itself to Tc<\/sub> by doing work on the piston.<\/li>\n
    3. c to d \u2014 isothermal compression at Tc<\/sub>.<\/strong> The gas touches the cold reservoir and is squeezed, rejecting Qc<\/sub> while holding at Tc<\/sub>.<\/li>\n
    4. d to a \u2014 adiabatic compression.<\/strong> Insulated again, the gas is compressed back to its starting volume and warms to Th<\/sub>. Ready to repeat.<\/li>\n<\/ol>\n \n

      The two isothermal legs obey Boyle’s law<\/a>, since pressure and volume trade off at fixed temperature. The two adiabatic legs are steeper, because the gas is losing internal energy as well as expanding \u2014 behaviour that falls straight out of the ideal gas law<\/a>.<\/p>\n \n

      \"Carnot<\/figure>\n \n

      The Carnot cycle on a pressure\u2013volume diagram. Drawn to scale for a diatomic ideal gas \u2014 the adiabats really are steeper than the isotherms.<\/p>\n \n

      Where the Formula Actually Comes From<\/h3>\n \n

      On the two isothermal legs the heat exchanged works out proportional to the reservoir temperature. For a reversible cycle that gives a strikingly clean result:<\/p>\n \n

      Qh \/ Th = Qc \/ Tc<\/div>\n \n

      Since the gas ends each cycle exactly where it started, its internal energy is unchanged, so W = Qh<\/sub> \u2212 Qc<\/sub>. Divide W by Qh<\/sub>, substitute the relation above, and Q drops out entirely \u2014 leaving 1 \u2212 Tc<\/sub>\/Th<\/sub>. If you want to see every integral, the University of Virginia keeps a full derivation of the Carnot cycle<\/a> online.<\/p>\n \n

      That cancellation is the whole story. The mass of gas, the pressure, the cylinder size \u2014 all gone. Only the two temperatures survive.<\/p>\n \n

      The Cleanest Picture: a Rectangle<\/h3>\n \n

      Plot the same cycle against entropy instead of volume and the shape becomes a perfect rectangle. Heat in is the top edge times its width; heat out is the bottom edge times the same width; work is the area in between.<\/p>\n \n

      \"Carnot<\/figure>\n \n

      On temperature\u2013entropy axes the Carnot cycle is a rectangle. Divide the shaded area by the area beneath the top edge and you get 1 \u2212 Tc<\/sub>\/Th<\/sub> in one line.<\/p>\n \n

      Look at the rectangle and the formula is almost visual. Work divided by heat in equals height divided by total height \u2014 which is exactly (Th<\/sub> \u2212 Tc<\/sub>)\/Th<\/sub>.<\/p>\n \n

      <\/span>Carnot Efficiency Lab<\/span><\/div>