Physics · Thermodynamics · NEET
In the Carnot cycle the heat is exchanged only during the two isothermal steps. Using the isothermal work formula, heat taken from the source is Q1 = nR T1 ln(V2/V1) and heat given to the sink is Q2 = nR T2 ln(V3/V4). For the two adiabatic steps, the relation T V^(gamma-1) = constant makes the volume ratios equal, so V2/V1 = V3/V4. The logarithm terms cancel, leaving Q2/Q1 = T2/T1. That is the key step that replaces heats with temperatures.
The formula eta = 1 - T2/T1 comes from an ideal gas relation where temperature is absolute (measured from 0 K). If you put celsius values, the ratio T2/T1 is wrong. Always convert first: T(K) = T(deg C) + 273. For example 100 deg C becomes 373 K, not 100.
eta = 1 - Q2/Q1 is the general definition for ANY heat engine (efficiency = work done / heat taken = (Q1 - Q2)/Q1). eta = 1 - T2/T1 is a SPECIAL result that is true only for a reversible Carnot engine, because only then Q2/Q1 equals T2/T1. For a real engine Q2/Q1 is larger than T2/T1, so its efficiency is lower.
No. The number of moles n and the gas constant R cancel out when we take the ratio Q2/Q1. Carnot's theorem states the efficiency of a Carnot engine is independent of the working substance. It depends only on the two temperatures T1 and T2.
Because efficiency is fraction of heat converted to work: eta = W/Q1 = (Q1 - Q2)/Q1 = 1 - Q2/Q1. The rejected (wasted) heat Q2 sits on top, so the smaller temperature T2 goes on top. A smaller T2/T1 ratio means less heat wasted and higher efficiency.
The efficiency of an ideal heat engine working between the freezing point and boiling point of water is:
A Carnot engine has an efficiency of 50% when its source is at 327 deg C. The temperature of the sink is:
A Carnot engine having an efficiency of 1/10 as a heat engine is used as a refrigerator. If the work done on the system is 10 J, the amount of energy absorbed from the reservoir at lower temperature is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
eta = 1 - T2/T1, where T1 is the source (hot) temperature and T2 is the sink (cold) temperature, both in kelvin. Equivalently eta = 1 - Q2/Q1 in terms of heat.
No. eta = 1 - T2/T1 equals 1 (100%) only if T2 = 0 K (absolute zero) or T1 is infinite. Both are impossible, so real efficiency is always below 100%. This links to the second law of thermodynamics.
Yes. A bigger gap between T1 and T2 (a hotter source or a colder sink) makes the ratio T2/T1 smaller, so eta = 1 - T2/T1 is larger.
No. By Carnot's theorem the efficiency depends only on T1 and T2, not on the type or amount of gas. The moles n and constant R cancel in the derivation.
Because it is reversible. No engine working between the same two temperatures can beat a Carnot engine. Any real (irreversible) engine wastes more heat and has lower efficiency.
