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PhysicsMediumClass 11

Entropy & Second Law — Heat pumps, refrigerators

Thermodynamics & KTG

21

JEE Qs

8%

Hard

75

min

Clearly visualize the direction of heat flow and work input for each device (refrigerator vs. heat pump) and always use absolute temperatures (Kelvin) in all formulas.

🧮 Key Formulas

Coefficient of Performance (Refrigerator): COP_R = Q_L / W_in = T_L / (T_H - T_L)
Coefficient of Performance (Heat Pump): COP_HP = Q_H / W_in = T_H / (T_H - T_L)
Relation between COPs: COP_HP = COP_R + 1
First Law for Refrigerator/Heat Pump: W_in = Q_H - Q_L
Clausius Statement of Second Law: Heat cannot flow spontaneously from a colder to a hotter body.
Kelvin-Planck Statement of Second Law: It is impossible to construct an engine that will work in a complete cycle and convert all heat supplied to it into mechanical work.
Entropy change for a reversible process: dS = dQ_rev / T
Entropy change for an isolated system (Second Law): Delta S_universe >= 0 (for any natural process)

✅ Key Points for JEE

  • 1Refrigerators and heat pumps are essentially heat engines working in reverse, requiring external work to transfer heat from a colder to a hotter body.
  • 2The Coefficient of Performance (COP) quantifies the efficiency of refrigerators and heat pumps, defined as the ratio of desired heat transfer to the work input.
  • 3For ideal (Carnot) refrigerators/heat pumps, COP depends only on the absolute temperatures of the hot (T_H) and cold (T_L) reservoirs.
  • 4Unlike efficiency (which is always < 1), COP can be greater than 1, meaning more heat can be moved than the work input.
  • 5The Second Law of Thermodynamics (Clausius and Kelvin-Planck statements) imposes fundamental limits on the operation of heat engines, refrigerators, and heat pumps, prohibiting perpetual motion machines of the second kind.

⚠️ Common Mistakes

  • Confusing the definitions of COP for refrigerators (Q_L/W) and heat pumps (Q_H/W).
  • Using Celsius instead of Kelvin for temperature values in COP formulas, leading to incorrect results.
  • Assuming COP is always less than 1, similar to thermal efficiency of heat engines.
  • Incorrectly identifying Q_H and Q_L for the system (e.g., Q_H is rejected by a refrigerator, but absorbed by a heat pump for heating).

📝 Practice Questions

See all

Q29.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

2025·Assertion ReasoningMedium

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

2025·NumericalMedium

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

2025·MCQHard

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

2025·MCQMedium

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

2025·MCQMedium

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

2025·NumericalMedium

NCERT Chapters

  • Class 11 Physics Ch 12: Thermodynamics