Heat Engines — Efficiency, Carnot engine
Thermodynamics & KTG
21
JEE Qs
8%
Hard
75
min
Always convert temperatures to Kelvin for efficiency calculations and clearly identify heat absorbed (Q_H) and heat rejected (Q_C) to avoid sign errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1A heat engine operates in a cyclic process, absorbing heat (Q_H) from a hot reservoir (source, T_H), converting a portion into mechanical work (W), and rejecting the remaining heat (Q_C) to a cold reservoir (sink, T_C).
- 2The efficiency (η) of any heat engine is defined as the ratio of net work done (W) by the engine to the heat absorbed from the hot reservoir (Q_H), i.e., η = W/Q_H.
- 3The Carnot engine is a theoretical, ideal, and reversible heat engine operating between two temperature reservoirs, achieving the maximum possible efficiency (η_Carnot = 1 - T_C/T_H) for those temperatures.
- 4According to Carnot's theorem, no engine operating between two given temperatures can be more efficient than a Carnot engine operating between the same two temperatures; all reversible engines operating between the same two temperatures have the same efficiency.
- 5The efficiency of any real or ideal heat engine is always less than 1 (or 100%) due to the Kelvin-Planck statement of the Second Law of Thermodynamics.
⚠️ Common Mistakes
- ✕Using Celsius temperatures instead of absolute (Kelvin) temperatures for T_H and T_C in the Carnot efficiency formula (η_Carnot = 1 - T_C/T_H).
- ✕Incorrectly interpreting Q_H and Q_C (heat absorbed from hot, heat rejected to cold) or confusing their signs when applying W = Q_H - Q_C (Q_H and Q_C usually denote magnitudes here).
- ✕Assuming that the Carnot efficiency formula (1 - T_C/T_H) applies to *any* heat engine, rather than exclusively to ideal reversible (Carnot) engines operating between those specific temperatures.
- ✕Forgetting that W in the formulas refers to the *net* work done by the engine over one complete cycle.
📝 Practice Questions
See allQ29.Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process. Reason (R) : In isothermal process, PV = constant, while in adiabatic process PVγ = constant. Here γ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas. In the light of the above statements, choose the correct answer from the options given below : (1) Both (A) and (R) are true and (R) is the correct (2) Both (A) and (R) are true but (R) is NOT the explanation of (A) correct explanation of (A) (3) (A) is true but (R) is false (4) (A) is false but (R) is true 2025 (29 Jan Shift 2) JEE Main Previous Year Paper
Q40.A cup of coffee cools from 90∘C to 80∘C in t minutes when the room temperature is 20∘C. The time taken by the similar cup of coffee to cool from 80∘C to 60∘C at the same room temperature is : (1) 13 10 t (2) 1013 t (3) 13 5 t (4) 135 t
Q42. A poly-atomic molecule ( CV = 3R, CP = 4R, where R is gas constant) goes from phase space point A (PA = 105 Pa, VA = 4 × 10−6 m3) to point B (PB = 5 × 104 Pa, VB = 6 × 10−6 m3) to point 2025 (29 Jan Shift 2) JEE Main Previous Year Paper C (PC = 104 Pa, VC = 8 × 10−6 m3). A to B is an adiabatic path and B to C is an isothermal path. The net heat absorbed per unit mole by the system is : (1) 500R(ln 3 + ln 4) (2) 450R(ln 4 −ln 3) (3) 500R ln 2 (4) 400R ln 4
Q29.An amount of ice of mass 10−3 kg and temperature −10∘C is transformed to vapour of temperature 110∘C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = 2100Jkg−1 K−1 , specific heat of water = 4180Jkg−1 K−1 , specific heat of steam = 1920Jkg−1 K−1 , Latent heat of ice = 3.35 × 105Jkg−1 and Latent heat of steam = 2.25 × 106 Jkg−1 ) (1) 3043 J (2) 3024 J (3) 3003 J (4) 3022 J
Q41.Two spherical bodies of same materials having radii 0.2 m and 0.8 m are placed in same atmosphere. The temperature of the smaller body is 800 K and temperature of the bigger body is 400 K . If the energy radiated from the smaller body is E, the energy radiated from the bigger body is (assume, effect of the surrounding temperature to be negligible), (1) 16 E (2) E (3) 64 E (4) 256 E
Q49.Three conductors of same length having thermal conductivity k1, k2 and k3 are connected as shown in figure. Area of cross sections of 1st and 2nd conductor are same and for 3rd conductor it is double of the 1st conductor. The temperatures are given in the figure. In steady state condition, the value of θ is _______ ∘C. (Given : k1 = 60Js−1 m−1 K−1, k2 = 120Js−1 m−1 K−1, k3 = 135Js−1 m−1 K−1 ) ∣ ∣∣ 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
NCERT Chapters
- Class 11 Physics Ch 12: Thermodynamics