K2KARAKORAM
मराठीEnglish
Open the app

State Board · Class 12 · Physics

Lesson 4: Thermodynamics

7 concepts · 56 practice questions with answers · Free to read

This lesson is also in: मराठी

Zeroth law, heat, internal energy, work and calorimetry

Link to earlier classes : The heat, temperature and energy conservation of classes 9-10 are extended here to the laws of thermodynamics.

From daily life

Mixing cold milk into hot tea gives a common middle temperature; the heat the tea loses is the heat the milk gains.

Simple definition

Heat is energy in transit because of a temperature difference. Internal energy = the total energy of the molecules. Calorimetry is the technique of measuring heat.

Hot and cold bodies in contact : both reach the final temperature T_fhotcoldT_ftTheat lost = heat gained

In the figure : The hot body's T falls, the cold body's rises; both meet at T_f.

Rule 1: Zeroth law — If A is in thermal equilibrium with B and B with C, then A is in thermal equilibrium with C. This gives the idea of temperature and the principle of a thermometer.

  • Easy : When a thermometer reaches equilibrium with a patient its reading = the patient's temperature.
  • Medium : There is no net heat flow between bodies at the same temperature.
  • Exam (CET) : Temperature is the property telling equilibrium; heat is energy in transit. (Options : 1) temperature = energy 2) heat = temperature 3) temperature ≠ heat 4) both same → answer 3)

Rule 2: Heat, specific heat and work — Q = m c ΔT (c = specific heat); phase change : Q = m L; work at constant pressure W = P ΔV. Internal energy U = total kinetic + potential energy of the molecules.

  • Easy : 0.5 kg of water 20 °C → 80 °C (c = 4200) : Q = 0.5 × 4200 × 60 = 126000 J.
  • Medium : Melting 0.2 kg ice at 0 °C (L = 3.34 × 10⁵ J/kg) : Q = 66800 J.
  • Exam (CET) : Volume rises by 2 L at P = 10⁵ Pa : W = 10⁵ × 2 × 10⁻³ = 200 J. (Options : 1) 20 J 2) 200 J 3) 2000 J 4) 0.2 J → answer 2)

Rule 3: Calorimetry — In an isolated system heat lost = heat gained (energy conservation). On mixing two bodies T_f = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂).

  • Easy : 200 g water at 80 °C + 300 g water at 20 °C : T_f = (200 × 80 + 300 × 20) ÷ 500 = 44 °C.
  • Medium : Equal masses of the same liquid at 60 °C and 20 °C : T_f = 40 °C (the mean).
  • Exam (CET) : 0.1 kg metal (c = 400) at 100 °C into 0.2 kg water (20 °C) : T_f = 23.64 °C. (Options : 1) 23.6 °C 2) 60 °C 3) 40 °C 4) 100 °C → answer 1)

Derivation : the formula for T_f

  • Step 1 : The hot body goes T₁ → T_f : heat lost Q₁ = m₁c₁ (T₁ − T_f).
  • Step 2 : The cold body goes T₂ → T_f : heat gained Q₂ = m₂c₂ (T_f − T₂).
  • Step 3 : Q₁ = Q₂ ⇒ m₁c₁T₁ − m₁c₁T_f = m₂c₂T_f − m₂c₂T₂.
  • Step 4 : Collecting T_f : T_f = (m₁c₁T₁ + m₂c₂T₂) ÷ (m₁c₁ + m₂c₂).
Remember

Q = m c ΔT; Q = m L; W = P ΔV; heat lost = heat gained; T_f = Σ(mcT) ÷ Σ(mc).

Common mistake / CET trap

Heat and temperature are not the same. During melting the temperature is constant yet heat is needed. Convert g to kg; check the units of c.

Shortcut

Memory clue : "Lost = gained" and equal masses of the same liquid ⇒ the mean temperature.

Spot the difference

QuantityFormulaUnit
Heatm c ΔTJ
Phase changem LJ
Work (const P)P ΔVJ
cQ ÷ (m ΔT)J kg⁻¹K⁻¹

Practice questions

Easy
  1. 1State the zeroth law.
  2. 2Unit of c in Q = m c ΔT?
  3. 3Heat to melt 1 kg of ice (L = 3.34 × 10⁵)?
Medium
  1. 4Heat 2 kg of water by 10 °C (c = 4200) : Q?
  2. 5P = 2 × 10⁵ Pa, ΔV = 3 × 10⁻³ m³ : W?
  3. 6100 g water at 90 °C + 100 g water at 30 °C : T_f?
Challenging
  1. 70.3 kg water at 70 °C + 0.2 kg water at 20 °C : T_f?
  2. 8Total Q to warm 1 kg ice from −10 °C to 0 °C and melt it fully?
Show answers (try them yourself first)
  1. 1A~B and B~C ⇒ A~C
  2. 2J kg⁻¹K⁻¹
  3. 3Q = 334000 J
  4. 4Q = 2 × 4200 × 10 = 84000 J
  5. 5W = 2 × 10⁵ × 3 × 10⁻³ = 600 J
  6. 6T_f = 60 °C
  7. 7T_f = (0.3 × 70 + 0.2 × 20) ÷ 0.5 = 50 °C
  8. 8Q = 1 × 2100 × 10 + 3.34 × 10⁵ = 355000 J

First law of thermodynamics and state variables

From daily life

A cycle pump gets warm when you compress air; the work you do raises the internal energy of the air.

Simple definition

First law : heat supplied to a system = increase in internal energy + work done by the system. It is the law of energy conservation.

First law : heat supplied Q = ΔU + WΔUQWW = P ΔV

In the figure : The gas under the piston receives Q; part raises ΔU and the rest pushes the piston up as W.

Rule 1: First law — Q = ΔU + W, i.e. ΔU = Q − W (energy conservation). Signs : Q gained by the system is positive; W done by the system is positive (expansion); W done on the system is negative (compression).

  • Easy : Q = 500 J, W = 200 J (by the system) ⇒ ΔU = 300 J.
  • Medium : A gas is compressed : W = −150 J and it loses heat Q = −50 J ⇒ ΔU = −50 − (−150) = +100 J.
  • Exam (CET) : Free expansion : Q = 0, W = 0 ⇒ ΔU = 0; the T of an ideal gas does not change. (Options : 1) T rises 2) T falls 3) T unchanged 4) P unchanged → answer 3)

Rule 2: State variables — P, V, T, U, n are state variables; their change depends only on the initial and final states. Q and W depend on the path. In a cyclic process ΔU = 0 ⇒ Q = W.

  • Easy : In one cycle Q = 400 J is absorbed ⇒ W = 400 J.
  • Medium : A → B by two different paths : ΔU is the same but Q and W differ.
  • Exam (CET) : Net W in a cycle = area of the P–V loop. (Options : 1) area of the loop 2) its length 3) P 4) V → answer 1)

Rule 3: U of an ideal gas — An ideal gas has no intermolecular forces so U depends only on T : ΔU = n C_V ΔT (for any process).

  • Easy : 1 mol monatomic gas T 300 → 400 K : ΔU = 1.5 × 8.314 × 100 = 1247 J.
  • Medium : In an isothermal process ΔU = 0 ⇒ Q = W.
  • Exam (CET) : ΔU between two states is independent of the path. (Options : 1) depends 2) does not depend 3) only on P 4) only on W → answer 2)

Derivation : derivation of W = P ΔV

  • Step 1 : A cylinder of cross-section A; the piston moves dx; force of the gas F = P A.
  • Step 2 : Work dW = F dx = P A dx.
  • Step 3 : A dx = dV (change in volume) ⇒ dW = P dV.
  • Step 4 : For constant P : W = P (V₂ − V₁); in general W = ∫ P dV = the area under the P–V graph.
Remember

ΔU = Q − W; W = ∫P dV; in a cycle ΔU = 0 ⇒ Q = W; ΔU = nC_VΔT.

Common mistake / CET trap

Sign of W : expansion positive, compression negative (work by the system). Q and W are not state variables. ΔU = n C_V ΔT holds for any process, not just constant V.

Shortcut

Memory clue : "Received Q = raised U + gave W" (Q = ΔU + W).

Spot the difference

QuantityTypeDepends on
P, V, T, Ustate variablestate only
Q, Wpath functionthe path
ΔUchange of a state variablenot the path

Practice questions

Easy
  1. 1Write the first law.
  2. 2ΔU in a cyclic process?
  3. 3Sign of W during expansion?
Medium
  1. 4Q = 300 J, W = 120 J : ΔU?
  2. 5200 J of work is done on a gas and it gives out 50 J of heat : ΔU?
  3. 6A P–V loop of area 400 J in a cycle : net Q?
Challenging
  1. 72 mol of monatomic gas T 300 → 350 K : ΔU?
  2. 8Why does the T of an ideal gas not change in free expansion?
Show answers (try them yourself first)
  1. 1Q = ΔU + W
  2. 2Zero
  3. 3Positive
  4. 4ΔU = 300 − 120 = 180 J
  5. 5ΔU = −50 − (−200) = 150 J
  6. 6Q = W = 400 J
  7. 7ΔU = 1.5 × 2 × 8.314 × 50 = 1247 J
  8. 8Q = 0, W = 0 ⇒ ΔU = 0; the U of an ideal gas depends only on T ⇒ T constant.

Work and heat in thermodynamic processes

From daily life

In a pressure cooker the volume is fixed and pressure rises; in a balloon the volume grows; each process has its own work and heat.

Simple definition

Process names : isothermal (T constant), isobaric (P constant), isochoric (V constant). Work = area under the P–V graph.

Isothermal, isobaric and isochoric processes on a P–V graphisothermal T constisobaric P constisochoricVP

In the figure : Three processes from one point : hyperbola, horizontal line and vertical line.

Live figure

Work and heat in thermodynamic processes

In the app you can move this figure with your finger.

Open in the app

Rule 1: Isothermal process — T constant ⇒ ΔU = 0 ⇒ Q = W = n R T ln (V₂ ÷ V₁). P V = constant. Graph : hyperbola.

  • Easy : 1 mol, 300 K, volume doubled : W = 8.314 × 300 × ln 2 = 1729 J.
  • Medium : In slow expansion or compression heat flows out/in and T stays constant.
  • Exam (CET) : In isothermal expansion the gas absorbs heat and does equal work. (Options : 1) U rises 2) Q = W 3) Q = 0 4) W = 0 → answer 2)

Rule 2: Isobaric process — P constant ⇒ W = P ΔV = n R ΔT, ΔU = n C_V ΔT, Q = n C_P ΔT. Graph : horizontal line.

  • Easy : 2 mol monatomic, T 300 → 400 K : Q = 2 × (5/2) × 8.314 × 100 = 4157 J; ΔU = 2494 J; W = 1663 J.
  • Medium : Q : ΔU : W = C_P : C_V : R. Monatomic ⇒ 5 : 3 : 2; diatomic ⇒ 7 : 5 : 2.
  • Exam (CET) : When a diatomic gas is heated at constant pressure the fraction of heat used for work = R ÷ C_P = 2/7 ≈ 28.6 %. (Options : 1) 40 % 2) 28.6 % 3) 60 % 4) 100 % → answer 2)

Rule 3: Isochoric and cyclic processes — V constant ⇒ W = 0 ⇒ Q = ΔU = n C_V ΔT. Cyclic : ΔU = 0, net W = area of the loop (clockwise ⇒ positive).

  • Easy : 1 mol diatomic, T 300 → 350 K, V constant : Q = 2.5 × 8.314 × 50 = 1039 J.
  • Medium : Rectangular loop : P 1 → 3 × 10⁵ Pa, V 1 → 3 × 10⁻³ m³ ⇒ W = 2 × 10⁵ × 2 × 10⁻³ = 400 J.
  • Exam (CET) : A cycle traversed anticlockwise ⇒ W negative (refrigerator). (Options : 1) positive 2) negative 3) zero 4) infinite → answer 2)

Derivation : isothermal work W = n R T ln (V₂ ÷ V₁)

  • Step 1 : W = ∫ P dV (from V₁ to V₂).
  • Step 2 : Ideal gas : P = n R T ÷ V with T constant.
  • Step 3 : W = n R T ∫ dV ÷ V = n R T [ln V] = n R T ln (V₂ ÷ V₁).
  • Step 4 : ΔU = 0 ⇒ Q = W = n R T ln (V₂ ÷ V₁).
Remember

Isothermal Q = W; isobaric W = PΔV, Q = nC_PΔT; isochoric W = 0, Q = ΔU; cycle ΔU = 0.

Common mistake / CET trap

Mixing ln with log₁₀ (ln x = 2.303 log₁₀ x). In isobaric work Q = n C_P ΔT, not n C_V ΔT. Check the direction when using the loop area.

Shortcut

Memory clue : "T fixed – Q = W; P fixed – Q : ΔU : W = 5 : 3 : 2 (monatomic)".

Spot the difference

ProcessConstantWΔUQ
IsothermalTnRT ln(V₂/V₁)0= W
IsobaricPPΔVnC_VΔTnC_PΔT
IsochoricV0nC_VΔT= ΔU

Practice questions

Easy
  1. 1ΔU in an isothermal process?
  2. 2W in an isochoric process?
  3. 3W in an isobaric process = ?
Medium
  1. 4P = 2 × 10⁵ Pa, V goes 1 L → 4 L : W?
  2. 51 mol at 300 K, V triples (isothermal) : W? (ln 3 = 1.0986)
  3. 61 mol monatomic gas, V constant, T rises 100 K : Q?
Challenging
  1. 7Monatomic gas at constant P : what percentage of Q becomes work?
  2. 81 mol diatomic, constant P, T up by 50 K : find Q, ΔU, W.
Show answers (try them yourself first)
  1. 1Zero
  2. 2Zero
  3. 3P ΔV
  4. 4W = 2 × 10⁵ × 3 × 10⁻³ = 600 J
  5. 5W = 8.314 × 300 × 1.0986 = 2740 J
  6. 6Q = 1.5 × 8.314 × 100 = 1247 J
  7. 7W ÷ Q = R ÷ C_P = 2 ÷ 5 = 40 %
  8. 8Q = 3.5 × 8.314 × 50 = 1455 J; ΔU = 1039 J; W = 416 J

Adiabatic process

From daily life

A spray can turns cold when gas rushes out; there is no time for heat to enter or leave — an adiabatic process.

Simple definition

An adiabatic process has no heat exchange (Q = 0). In it P V^γ = constant.

The adiabatic curve is steeper than the isothermal (slope γ times)isothermaladiabaticVP

In the figure : From the same point the adiabatic (red) falls more steeply than the isothermal (blue).

Rule 1: Adiabatic process — Q = 0 (no heat exchange). P V^γ = constant, T V^(γ−1) = constant, T^γ P^(1−γ) = constant. ΔU = −W.

  • Easy : Air suddenly compressed in a pump : Q ≈ 0, work is done on the gas ⇒ T rises.
  • Medium : Diatomic (γ = 1.4), V halved, T₁ = 300 K : T₂ = 300 × 2^0.4 = 396 K; P₂ = 2.64 P₁.
  • Exam (CET) : Rapid expansion ⇒ the gas cools (a spray can feels cold). (Options : 1) T rises 2) T falls 3) T same 4) Q rises → answer 2)

Rule 2: Adiabatic work — W = (P₁V₁ − P₂V₂) ÷ (γ − 1) = n R (T₁ − T₂) ÷ (γ − 1). In expansion W is positive and T falls; in compression W is negative and T rises.

  • Easy : The compression above (1 mol) : W = 8.314 × (300 − 395.9) ÷ 0.4 = -1992 J (done on the gas).
  • Medium : ΔU = n C_V ΔT = −W (Q = 0).
  • Exam (CET) : For the same volume change the adiabatic curve is steeper, so the adiabatic work is smaller. (Options : 1) more 2) less 3) same 4) zero → answer 2 for expansion)

Rule 3: Adiabatic vs isothermal — Isothermal : P V = constant (slope −P ÷ V). Adiabatic : P V^γ = constant (slope −γ P ÷ V). So at a given point adiabatic slope = γ × isothermal slope.

  • Easy : γ = 5/3 : adiabatic slope = 1.67 × isothermal.
  • Medium : In isothermal compression heat flows out to keep T fixed; in adiabatic compression T rises.
  • Exam (CET) : A larger γ = C_P ÷ C_V makes the adiabatic curve steeper. (Options : 1) less steep 2) steeper 3) no change 4) straight → answer 2)

Derivation : P V^γ = constant

  • Step 1 : Q = 0 ⇒ 0 = dU + P dV ⇒ n C_V dT = − P dV.
  • Step 2 : Differentiate P V = n R T : P dV + V dP = n R dT.
  • Step 3 : Eliminate dT using the first equation and C_P = C_V + R : C_V (P dV + V dP) = − R P dV.
  • Step 4 : Simplifying C_V V dP + C_P P dV = 0 ⇒ dP ÷ P + γ dV ÷ V = 0.
  • Step 5 : Integrating : ln P + γ ln V = constant ⇒ P V^γ = constant.
Remember

Q = 0; P V^γ = constant; T V^(γ−1) = constant; W = nR(T₁ − T₂) ÷ (γ − 1); ΔU = −W.

Common mistake / CET trap

In an adiabatic process T is not constant, and although Q = 0, ΔU is not zero. The value of γ depends on the gas (f = 3 → 5/3, f = 5 → 7/5).

Shortcut

Memory clue : "Adiabatic = Q zero, slope γ times" and "squeeze – hot, spread – cold".

Spot the difference

PointIsothermalAdiabatic
ConstantTQ = 0
EquationP V = constantP V^γ = constant
Slope of curve−P ÷ V−γ P ÷ V
T on compressionconstantrises

Practice questions

Easy
  1. 1Q in an adiabatic process?
  2. 2For which process is P V^γ constant?
  3. 3Why does a gas cool on rapid expansion?
Medium
  1. 4γ = 1.4, V₂ = V₁ ÷ 2, P₁ = 1 atm : P₂? (2^1.4 = 2.639)
  2. 5Adiabatic : ΔU = 200 J ⇒ W?
  3. 61 mol, γ = 5/3, T falls 400 → 300 K (expansion) : W?
Challenging
  1. 7A monatomic gas is compressed adiabatically to V ÷ 8, T₁ = 300 K : T₂? (8^(2/3) = 4)
  2. 8Ratio of adiabatic to isothermal slope for a diatomic gas?
Show answers (try them yourself first)
  1. 1Zero
  2. 2Adiabatic
  3. 3It does work at the expense of internal energy, Q = 0
  4. 4P₂ = 1 × 2.639 = 2.64 atm
  5. 5W = −200 J (done on the gas)
  6. 6W = 8.314 × 100 ÷ (2/3) = 1247 J
  7. 7T₂ = 300 × 4 = 1200 K
  8. 8γ = 1.4

Heat engines and efficiency

From daily life

A motorcycle engine burns petrol to turn the wheels, but much of the heat leaves through the silencer.

Simple definition

A heat engine converts heat into work. It takes Q_H from a hot reservoir, gives W and rejects Q_C to a cold reservoir.

Heat engine : absorbs Q_H, gives W, rejects Q_Chot reservoir T_Hcold reservoir T_CengineQ_HQ_CW

In the figure : Q_H in from above, W out to the right, Q_C out below.

Rule 1: Heat engine and efficiency — In a cycle W = Q_H − Q_C. Efficiency η = W ÷ Q_H = 1 − Q_C ÷ Q_H. η is always less than 1 (100 % is impossible).

  • Easy : Q_H = 1000 J, Q_C = 600 J ⇒ W = 400 J; η = 40 %.
  • Medium : η = 25 %, W = 500 J ⇒ Q_H = 500 ÷ 0.25 = 2000 J; Q_C = 1500 J.
  • Exam (CET) : An engine takes in 800 J and rejects 600 J : η = 200 ÷ 800 = 25 %. (Options : 1) 75 % 2) 25 % 3) 33 % 4) 12.5 % → answer 2)

Rule 2: Types of engines — External combustion (steam engine) : fuel burns outside the cylinder. Internal combustion (petrol/diesel) : inside the cylinder. In both Q_H → W + Q_C.

  • Easy : Compare a steam engine and a motorcycle engine.
  • Medium : An internal combustion engine has an efficiency of roughly 25–40 %.
  • Exam (CET) : The remaining heat goes to the surroundings, hence η < 100 %. (Options : 1) friction only 2) Q_C 3) Q_H 4) W → answer 2)

Rule 3: The engine cycle — In each cycle ΔU = 0 ⇒ net work = area of the P–V loop = Q_H − Q_C. A bigger loop ⇒ more work.

  • Easy : Loop area 250 J ⇒ W = 250 J per cycle.
  • Medium : 20 cycles per second at 250 J each ⇒ power = 5000 W.
  • Exam (CET) : Power P = W × (cycles ÷ s). (Options : 1) 5 kW 2) 50 W 3) 500 W 4) 12.5 W → answer 1)

Derivation : the formula for η

  • Step 1 : In one cycle ΔU = 0 ⇒ first law : Q_net = W.
  • Step 2 : Q_net = Q_H − Q_C (absorbed − rejected) ⇒ W = Q_H − Q_C.
  • Step 3 : Efficiency = work obtained ÷ heat supplied = W ÷ Q_H.
  • Step 4 : Substituting W : η = 1 − Q_C ÷ Q_H.
Remember

W = Q_H − Q_C; η = W ÷ Q_H = 1 − Q_C ÷ Q_H; power = W × cycles/s.

Common mistake / CET trap

Multiply by 100 for percent. Q_C is not lost, it is rejected to the surroundings. Do not take η = Q_C ÷ Q_H.

Shortcut

Memory clue : "Work ÷ heat taken" – η is always W over Q_H.

Spot the difference

QuantityFormulaUnit
WorkW = Q_H − Q_CJ
Efficiencyη = W ÷ Q_Hno unit
PowerW × cycles/sW (watt)

Practice questions

Easy
  1. 1Write the formula for η.
  2. 2ΔU in a cycle?
  3. 3Is η = 100 % possible?
Medium
  1. 4Q_H = 500 J, Q_C = 350 J : W and η?
  2. 5η = 30 %, W = 300 J : Q_H?
  3. 6An engine takes 2000 J and rejects 1500 J : η?
Challenging
  1. 7An engine does 40 J per cycle at 50 cycles/s : power?
  2. 8Q_C ÷ Q_H for an engine of η = 20 %?
Show answers (try them yourself first)
  1. 1η = W ÷ Q_H
  2. 2Zero
  3. 3No; Q_C cannot be zero
  4. 4W = 150 J; η = 30 %
  5. 5Q_H = 300 ÷ 0.3 = 1000 J
  6. 6η = 500 ÷ 2000 = 25 %
  7. 7P = 40 × 50 = 2000 W
  8. 8Q_C ÷ Q_H = 1 − 0.2 = 0.8

Refrigerators and heat pumps

From daily life

The coils behind a fridge feel warm; heat from inside is pushed out with the help of W.

Simple definition

A refrigerator is a reversed heat engine : using W it pulls heat from the cold side and dumps it on the hot side. COP = benefit ÷ cost.

Refrigerator : W moves Q_C out and Q_H is dumpedroom T_Hinside T_CunitQ_CQ_HW

In the figure : Q_C in from below, W in, Q_H out at the top.

Rule 1: Refrigerator and COP — In a cycle Q_H = Q_C + W. Coefficient of performance COP (β) = Q_C ÷ W = Q_C ÷ (Q_H − Q_C). β can be greater than 1.

  • Easy : Q_C = 300 J, W = 100 J ⇒ Q_H = 400 J; β = 3.
  • Medium : β = 4, W = 50 J ⇒ Q_C = 200 J.
  • Exam (CET) : Q_H = 500 J, Q_C = 400 J : β = 400 ÷ 100 = 4. (Options : 1) 5 2) 4 3) 0.8 4) 0.25 → answer 2)

Rule 2: Heat pump — A heat pump warms a room by delivering Q_H inside. β_hp = Q_H ÷ W = β + 1.

  • Easy : The same device used as a heat pump : β_hp = 400 ÷ 100 = 4.
  • Medium : β = 5 ⇒ β_hp = 6.
  • Exam (CET) : β_hp is always 1 more than β. (Options : 1) 1 less 2) 1 more 3) same 4) double → answer 2)

Rule 3: Carnot refrigerator — For an ideal refrigerator β = T_C ÷ (T_H − T_C) (kelvin). Lower T_C or larger difference ⇒ smaller β.

  • Easy : T_C = 270 K, T_H = 300 K ⇒ β = 270 ÷ 30 = 9.
  • Medium : A freezer (255 K) has smaller β than a fridge (270 K).
  • Exam (CET) : Do not use °C; add 273. (Options : 1) use °C 2) use K 3) both work 4) F → answer 2)

Derivation : the formula for β

  • Step 1 : In a cycle ΔU = 0 ⇒ Q_C − Q_H + W = 0 (W given in).
  • Step 2 : So Q_H = Q_C + W ⇒ W = Q_H − Q_C.
  • Step 3 : Benefit = Q_C removed from the cold side; cost = W.
  • Step 4 : β = Q_C ÷ W = Q_C ÷ (Q_H − Q_C); for a heat pump the benefit is Q_H ⇒ β_hp = Q_H ÷ W = β + 1.
Remember

β = Q_C ÷ W; Q_H = Q_C + W; β_hp = β + 1; Carnot β = T_C ÷ (T_H − T_C).

Common mistake / CET trap

Q_H = Q_C + W (not a difference). Use kelvin in the Carnot formula. Call β a coefficient; β > 1 is fine.

Shortcut

Memory clue : "Fridge = cold is the gain, pump = heat is the gain".

Spot the difference

DeviceBenefitCOP
RefrigeratorQ_CQ_C ÷ W
Heat pumpQ_HQ_H ÷ W
Carnot refrig.T_CT_C ÷ (T_H − T_C)

Practice questions

Easy
  1. 1What is the COP of a refrigerator?
  2. 2Relation among Q_H, Q_C and W?
  3. 3β_hp = ?
Medium
  1. 4Q_C = 600 J, W = 150 J : β?
  2. 5β = 3, Q_C = 450 J : W and Q_H?
  3. 6T_C = 260 K, T_H = 300 K : β?
Challenging
  1. 7Carnot β between −3 °C and 27 °C?
  2. 8β_hp of a refrigerator with β = 6?
Show answers (try them yourself first)
  1. 1β = Q_C ÷ W
  2. 2Q_H = Q_C + W
  3. 3β + 1
  4. 4β = 600 ÷ 150 = 4
  5. 5W = 150 J; Q_H = 600 J
  6. 6β = 260 ÷ 40 = 6.5
  7. 7β = 270 ÷ 30 = 9
  8. 8β_hp = 7

Second law and the Carnot cycle

From daily life

Hot tea cools by itself, but cold tea never warms by itself; heat flows in one direction.

Simple definition

The second law sets the direction of heat flow and conversion. The Carnot engine is the most efficient ideal engine between two temperatures.

Carnot cycle : two isothermals and two adiabaticsABCDAB : T_H (Q_H absorbed)CD : T_C (Q_C rejected)BC adiabaticDA adiabaticVP

In the figure : The area of the loop A→B→C→D = net work.

Rule 1: Second law — Kelvin–Planck : no cyclic engine can take heat from one reservoir and turn all of it into work (η = 100 % is impossible). Clausius : heat cannot by itself flow from cold to hot without work.

  • Easy : A fridge needs electricity (W) to move heat out; it will not do so by itself.
  • Medium : A ship cannot run by drawing heat only from the sea (one reservoir).
  • Exam (CET) : The two statements are equivalent. (Options : 1) different 2) equivalent 3) opposite 4) both wrong → answer 2)

Rule 2: Carnot cycle and efficiency — Four steps : AB isothermal expansion (Q_H), BC adiabatic expansion, CD isothermal compression (Q_C), DA adiabatic compression. Q_C ÷ Q_H = T_C ÷ T_H ⇒ η = 1 − T_C ÷ T_H.

  • Easy : T_H = 600 K, T_C = 300 K ⇒ η = 1 − 0.5 = 50 %.
  • Medium : Q_H = 2000 J, η = 50 % ⇒ W = 1000 J; Q_C = 1000 J.
  • Exam (CET) : 127 °C and 27 °C : T_H = 400 K, T_C = 300 K ⇒ η = 25 %. (Options : 1) 21 % 2) 25 % 3) 75 % 4) 79 % → answer 2)

Rule 3: Carnot theorem and limits — Between given T_H and T_C no engine is more efficient than a Carnot engine. All reversible engines have the same η. η = 1 only if T_C = 0 K (impossible).

  • Easy : Carnot η is 40 %; another engine claims 45 % ⇒ impossible.
  • Medium : To raise η increase T_H or decrease T_C.
  • Exam (CET) : Maximum η between 500 K and 300 K = 40 %. (Options : 1) 60 % 2) 40 % 3) 80 % 4) 100 % → answer 2)

Derivation : Carnot η = 1 − T_C ÷ T_H

  • Step 1 : AB (isothermal T_H) : Q_H = n R T_H ln (V_B ÷ V_A).
  • Step 2 : CD (isothermal T_C) : Q_C = n R T_C ln (V_C ÷ V_D).
  • Step 3 : Adiabatic BC and DA : T V^(γ−1) constant ⇒ V_B ÷ V_A = V_C ÷ V_D.
  • Step 4 : The ln factors are equal ⇒ Q_C ÷ Q_H = T_C ÷ T_H.
  • Step 5 : η = 1 − Q_C ÷ Q_H = 1 − T_C ÷ T_H.
Remember

η = 1 − T_C ÷ T_H; Q_C ÷ Q_H = T_C ÷ T_H; Carnot β = T_C ÷ (T_H − T_C).

Common mistake / CET trap

T in kelvin, not °C. η is never 100 %. Q_C ÷ Q_H = T_C ÷ T_H holds only for a Carnot/reversible engine.

Shortcut

Memory clue : "Hotter source, colder sink, higher η".

Spot the difference

QuantityFormulaUnit
Carnot η1 − T_C ÷ T_Hno unit
Q ratioQ_C ÷ Q_H = T_C ÷ T_H—
Carnot βT_C ÷ (T_H − T_C)no unit
TemperatureK = °C + 273K

Practice questions

Easy
  1. 1How many processes does a Carnot cycle have, and which?
  2. 2State the Kelvin–Planck statement.
  3. 3Formula for Carnot η?
Medium
  1. 4T_H = 800 K, T_C = 400 K : η?
  2. 5T_H = 327 °C, T_C = 27 °C : η?
  3. 6Q_H = 1200 J, T_C ÷ T_H = 0.25 : W?
Challenging
  1. 7With T_C = 300 K, what T_H gives η = 50 %?
  2. 8An engine works between 500 K and 300 K, takes 900 J and rejects 500 J. Is it possible?
Show answers (try them yourself first)
  1. 1Four : two isothermal and two adiabatic
  2. 2A cycle converting heat from one reservoir entirely into work is impossible
  3. 3η = 1 − T_C ÷ T_H
  4. 4η = 1 − 0.5 = 50 %
  5. 5T_H = 600 K, T_C = 300 K ⇒ η = 50 %
  6. 6W = 1200 × 0.75 = 900 J
  7. 7T_H = 300 ÷ 0.5 = 600 K
  8. 8η = 400 ÷ 900 = 44.4 % > 40 % (maximum) ⇒ impossible

Quick revision of the whole lesson

ConceptKey formula / factUnit / dimensions
HeatQ = m c ΔT; Q = m LJ; c : [L² T⁻² K⁻¹]
First lawQ = ΔU + W; W = ∫P dVJ
Ideal-gas UΔU = n C_V ΔTJ
IsothermalQ = W = nRT ln(V₂/V₁)J
AdiabaticP V^γ = constant; W = nR(T₁−T₂) ÷ (γ−1)γ has no unit
Heat engineη = W ÷ Q_H = 1 − Q_C ÷ Q_Hno unit
Refrigeratorβ = Q_C ÷ W; β_hp = β + 1no unit
Carnotη = 1 − T_C ÷ T_H; β = T_C ÷ (T_H−T_C)T in kelvin

🎯 Most important points for the exam

  • ΔU = Q − W (W done by the system); in a cycle ΔU = 0 ⇒ Q = W.
  • U depends only on T (ideal gas); ΔU = nC_VΔT in any process.
  • Isothermal Q = W; isobaric Q : ΔU : W = C_P : C_V : R; isochoric W = 0.
  • Adiabatic : Q = 0, PV^γ = constant, slope γ times.
  • η = 1 − Q_C ÷ Q_H; never 100 %.
  • β = Q_C ÷ W; heat pump β_hp = β + 1.
  • Carnot η = 1 − T_C ÷ T_H (kelvin); the most efficient.
Study mantra

Fix the sign → identify the process → right formula → T in kelvin → check the unit.