What is the efficiency of a Carnot engine?

η = 1 − T₂/T₁, where T₁ is the absolute temperature of the source and T₂ is the absolute temperature of the sink. No real engine can exceed Carnot efficiency.

Distinguish between isothermal and adiabatic processes.

In an isothermal process, temperature remains constant (ΔT = 0) and the gas exchanges heat with surroundings. In an adiabatic process, no heat is exchanged (ΔQ = 0) and the gas is thermally insulated. Isothermal: PV = constant; Adiabatic: PVᵞ = constant.

Entropy (S) is a measure of disorder in a system. The change in entropy is ΔS = ΔQ/T for a reversible process at temperature T. The second law states that the total entropy of an isolated system can only increase or remain constant.

Heat & Thermodynamics Exercise Solutions | KalviGavani

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How to Use This Page

1. Solve first, check later — Attempt every exercise question on your own before reading the solutions below. 2. Follow the sequence — Work through sections I to VI in order; later sections build on earlier ones. 3. Understand the method — Focus on the reasoning and formula selection, not just the final answer. 4. Practise numericals — Re-work the worked examples with different values until you can set them up without hints. 5. Review mistakes — Read the "Common Mistakes to Avoid" section before every exam.

Key Concepts (Quick Recap)

Temperature Scales

Scale	Freezing Point	Boiling Point	Relation
Celsius (°C)	0	100	C/5 = (F − 32)/9
Fahrenheit (°F)	32	212	F = (9C/5) + 32
Kelvin (K)	273.15	373.15	K = C + 273.15
Absolute zero	−273.15°C	−459.67°F	0 K

Thermal Expansion

Type	Formula	Coefficient
Linear expansion	ΔL = L₀αΔT	α (per °C or per K)
Area expansion	ΔA = A₀βΔT	β ≈ 2α
Volume expansion	ΔV = V₀γΔT	γ ≈ 3α
Relation	γ = 3α, β = 2α	—

Calorimetry and Specific Heat

Quantity	Formula	SI Unit
Heat exchanged	Q = mcΔT	J
Specific heat capacity	c = Q/(mΔT)	J/(kg·K)
Molar specific heat	C = Q/(nΔT)	J/(mol·K)
Latent heat (fusion)	Q = mL_f	J
Latent heat (vaporisation)	Q = mL_v	J
Heat lost = Heat gained	m₁c₁(T₁ − T) = m₂c₂(T − T₂)	—

First Law of Thermodynamics

Quantity	Formula	Sign Convention
First law	ΔQ = ΔU + ΔW	ΔQ > 0: heat absorbed
Work done by gas	ΔW = ∫P dV	ΔW > 0: gas expands
Internal energy	ΔU = nCᵥΔT	ΔU > 0: temperature rises

Thermodynamic Processes

Process	Condition	Work Done	Relation
Isothermal	ΔT = 0	W = nRT ln(V₂/V₁)	PV = constant
Adiabatic	ΔQ = 0	W = (P₁V₁ − P₂V₂)/(γ − 1)	PVᵞ = constant
Isobaric	P = constant	W = PΔV = nRΔT	V/T = constant
Isochoric	V = constant	W = 0	P/T = constant

Carnot Engine and Second Law

Quantity	Formula
Carnot efficiency	η = 1 − T₂/T₁
Carnot COP (refrigerator)	β = T₂/(T₁ − T₂)
Entropy change (reversible)	ΔS = ΔQ/T
Second law (Kelvin)	No engine can convert all heat into work
Second law (Clausius)	Heat cannot flow from cold to hot on its own

I. Choose the Best Answer (MCQs)

1. The temperature at which Celsius and Fahrenheit scales read the same is:

(a) −40°
(b) 0°
(c) −100°
(d) 40°

Answer(a) −40°

ExplanationSet C = F in the conversion formula: F = (9C/5) + 32. If C = F, then C = (9C/5) + 32, which gives C − 9C/5 = 32, so −4C/5 = 32, C = −40.

2. In an isothermal process, the change in internal energy of an ideal gas is:

(a) Positive
(b) Negative
(c) Zero
(d) Depends on volume

Answer(c) Zero

ExplanationFor an ideal gas, internal energy depends only on temperature (U = nCᵥT). In an isothermal process, temperature is constant (ΔT = 0), so ΔU = nCᵥ(0) = 0.

3. The work done in an isochoric process is:

(a) nRΔT
(b) PΔV
(c) Zero
(d) nRT ln(V₂/V₁)

Answer(c) Zero

ExplanationIn an isochoric process, volume remains constant (ΔV = 0). Since W = ∫P dV, when V is constant, W = 0.

4. The efficiency of a Carnot engine operating between 500 K and 300 K is:

(a) 20%
(b) 40%
(c) 60%
(d) 80%

Answer(b) 40%

Explanationη = 1 − T₂/T₁ = 1 − 300/500 = 1 − 0.6 = 0.4 = 40%.

5. Which of the following is a path function (not a state function)?

(a) Internal energy
(b) Entropy
(c) Enthalpy
(d) Work

Answer(d) Work

ExplanationWork depends on the path taken between initial and final states (e.g., isothermal work differs from adiabatic work between the same two states). Internal energy, entropy, and enthalpy are state functions — they depend only on the current state, not the path.

6. If 100 J of heat is supplied to a system and 40 J of work is done by the system, the change in internal energy is:

(a) 140 J
(b) 60 J
(c) −60 J
(d) −140 J

Answer(b) 60 J

ExplanationBy the first law, ΔQ = ΔU + ΔW. So ΔU = ΔQ − ΔW = 100 − 40 = 60 J.

7. The coefficient of volume expansion of a solid is related to its coefficient of linear expansion by:

(a) γ = α
(b) γ = 2α
(c) γ = 3α
(d) γ = α/3

Answer(c) γ = 3α

ExplanationFor an isotropic solid, volume expansion coefficient is approximately three times the linear expansion coefficient: γ = 3α. Similarly, area expansion coefficient β ≈ 2α.

II. Fill in the Blanks

1. The SI unit of specific heat capacity is J/(kg·K). 2. In an adiabatic process, ΔQ = 0 (no heat is exchanged with the surroundings). 3. The relationship between Celsius and Kelvin scales is K = °C + 273.15. 4. For an ideal gas in an isothermal process, PV = constant. 5. The coefficient of linear expansion has the SI unit K⁻¹ (or °C⁻¹). 6. The Carnot cycle consists of two isothermal and two adiabatic processes. 7. In the first law ΔQ = ΔU + ΔW, if the gas expands, ΔW is positive. 8. Entropy is a state function. 9. The latent heat of fusion of ice is approximately 3.36 × 10⁵ J/kg (or 336 kJ/kg). 10. The ratio of specific heats γ = Cₚ/Cᵥ for a monatomic ideal gas is 5/3 (≈ 1.67). 11. According to the second law of thermodynamics, the total entropy of an isolated system always increases. 12. The work done by a gas in isothermal expansion from V₁ to V₂ is nRT ln(V₂/V₁).

III. True or False (with corrections)

AnswerFalse — In an adiabatic process, the temperature of the gas remains constant. Correction: In an adiabatic process, no heat is exchanged (ΔQ = 0), but the temperature can change. During adiabatic expansion, the gas cools; during adiabatic compression, it heats up.

AnswerTrue — The efficiency of a Carnot engine depends only on the temperatures of the source and sink.

AnswerFalse — Work done is a state function. Correction: Work done is a path function. It depends on how the system moves from one state to another, not just on the initial and final states.

AnswerFalse — Latent heat is the heat required to change the temperature of a substance by 1°C. Correction: Latent heat is the heat required to change the state of a substance at constant temperature (e.g., melting or boiling). Specific heat capacity is the heat required to change the temperature by 1°C.

AnswerTrue — The coefficient of volume expansion of an ideal gas at constant pressure is 1/273 per degree Celsius (approximately).

IV. Match the Following

Column A	Column B
1. Isothermal process	(c) Temperature remains constant
2. Adiabatic process	(e) No heat exchange with surroundings
3. Isobaric process	(a) Pressure remains constant
4. Isochoric process	(f) Volume remains constant
5. Carnot cycle	(b) Most efficient cycle between two temperatures
6. Entropy	(d) Measure of disorder; always increases in isolated system

V. Answer Briefly (2-Mark Questions)

1. Distinguish between specific heat capacity and latent heat.

Answer

Specific heat capacity (c) is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K (unit: J/(kg·K)). Latent heat (L) is the amount of heat required to change the state of 1 kg of a substance at constant temperature (unit: J/kg). Specific heat involves temperature change; latent heat involves phase change at constant temperature.

2. What is a Carnot engine? State its efficiency formula.

Answer

A Carnot engine is an idealised heat engine that operates on the Carnot cycle — two isothermal and two adiabatic processes. It represents the maximum possible efficiency for any heat engine operating between two temperatures. Its efficiency is η = 1 − T₂/T₁, where T₁ is the source temperature and T₂ is the sink temperature (both in Kelvin).

3. State the first law of thermodynamics and explain the sign convention.

Answer

The first law states: ΔQ = ΔU + ΔW, where ΔQ is the heat supplied to the system, ΔU is the change in internal energy, and ΔW is the work done by the system. Sign convention: ΔQ is positive when heat is absorbed by the system; ΔW is positive when work is done by the system (gas expands); ΔU is positive when internal energy increases (temperature rises).

4. Define coefficient of linear expansion. Write its SI unit.

Answer

The coefficient of linear expansion (α) is defined as the fractional change in length per unit change in temperature: α = (ΔL/L₀)/ΔT. It represents how much a material expands per degree of temperature rise per unit original length. Its SI unit is K⁻¹ (or °C⁻¹, since a change of 1°C equals a change of 1 K). Typical values: steel ≈ 12 × 10⁻⁶ K⁻¹, aluminium ≈ 23 × 10⁻⁶ K⁻¹.

VI. Answer in Detail (5-Mark Questions)

1. Derive the relation between Cₚ and Cᵥ for an ideal gas (Mayer's relation) and solve a numerical.

For an ideal gas, the molar specific heat at constant pressure (Cₚ) exceeds the molar specific heat at constant volume (Cᵥ) because at constant pressure, the gas must do additional work (PΔV) as it expands.

At constant volume: ΔQ = nCᵥΔT = ΔU (since W = 0) At constant pressure: ΔQ = nCₚΔT = ΔU + ΔW

For an ideal gas: ΔU = nCᵥΔT (internal energy depends only on temperature) Work done at constant pressure: ΔW = PΔV = nRΔT (from PV = nRT)

Therefore:

nCₚΔT = nCᵥΔT + nRΔT

Dividing by nΔT:
Cₚ = Cᵥ + R   (Mayer's relation)

Also: γ = Cₚ/Cᵥ

For a monatomic gas:  Cᵥ = (3/2)R, Cₚ = (5/2)R, γ = 5/3 ≈ 1.67
For a diatomic gas:   Cᵥ = (5/2)R, Cₚ = (7/2)R, γ = 7/5 = 1.4

Worked Example: A diatomic ideal gas at 300 K is heated at constant pressure until its volume doubles. Find the heat supplied, change in internal energy, and work done for 2 moles of the gas. (R = 8.31 J/(mol·K))

Given:
  T₁ = 300 K, V₂ = 2V₁ (volume doubles), n = 2 mol
  Diatomic gas: Cᵥ = (5/2)R, Cₚ = (7/2)R, γ = 7/5

Step 1: Find final temperature
  At constant pressure, V/T = constant (Charles' law)
  V₁/T₁ = V₂/T₂ → T₂ = T₁ × (V₂/V₁) = 300 × 2 = 600 K

Step 2: Calculate ΔT
  ΔT = T₂ − T₁ = 600 − 300 = 300 K

Step 3: Calculate heat supplied (at constant pressure)
  ΔQ = nCₚΔT = 2 × (7/2) × 8.31 × 300
     = 2 × 3.5 × 8.31 × 300
     = 2 × 8725.5
     = 17451 J ≈ 17.45 kJ

Step 4: Calculate change in internal energy
  ΔU = nCᵥΔT = 2 × (5/2) × 8.31 × 300
     = 2 × 2.5 × 8.31 × 300
     = 2 × 6232.5
     = 12465 J ≈ 12.47 kJ

Step 5: Calculate work done
  ΔW = ΔQ − ΔU = 17451 − 12465 = 4986 J ≈ 4.99 kJ

  Verification: ΔW = nRΔT = 2 × 8.31 × 300 = 4986 J ✓

Answer: ΔQ ≈ 17.45 kJ, ΔU ≈ 12.47 kJ, ΔW ≈ 4.99 kJ

2. Describe the Carnot cycle with a PV diagram and derive its efficiency.

The Carnot cycle consists of four reversible processes:

1. Isothermal expansion (A → B): The gas absorbs heat Q₁ from the hot reservoir at temperature T₁ and expands isothermally. Work is done by the gas.

2. Adiabatic expansion (B → C): The gas expands adiabatically (no heat exchange). Temperature drops from T₁ to T₂.

3. Isothermal compression (C → D): The gas releases heat Q₂ to the cold reservoir at temperature T₂ and is compressed isothermally. Work is done on the gas.

4. Adiabatic compression (D → A): The gas is compressed adiabatically. Temperature rises from T₂ back to T₁.

Derivation of efficiency:

Heat absorbed from source: Q₁ (positive)
Heat rejected to sink:     Q₂ (positive quantity)

Efficiency: η = W/Q₁ = (Q₁ − Q₂)/Q₁ = 1 − Q₂/Q₁

For isothermal processes:
  Q₁ = nRT₁ ln(V₂/V₁)     [A → B, isothermal at T₁]
  Q₂ = nRT₂ ln(V₃/V₄)     [C → D, isothermal at T₂]

From adiabatic relations:
  T₁V₂^(γ−1) = T₂V₃^(γ−1)  [B → C]
  T₁V₁^(γ−1) = T₂V₄^(γ−1)  [D → A]

Dividing these two equations:
  (V₂/V₁)^(γ−1) = (V₃/V₄)^(γ−1)
  Therefore: V₂/V₁ = V₃/V₄

Substituting into Q₂/Q₁:
  Q₂/Q₁ = (nRT₂ ln(V₃/V₄)) / (nRT₁ ln(V₂/V₁))
         = T₂/T₁ × (ln(V₃/V₄)) / (ln(V₂/V₁))
         = T₂/T₁ × 1     [since V₂/V₁ = V₃/V₄]
         = T₂/T₁

Therefore: η = 1 − T₂/T₁

Worked Example: A Carnot engine operates between a source at 600 K and a sink at 300 K. If it absorbs 2000 J of heat per cycle, find the efficiency, work done per cycle, and heat rejected.

Given:
  T₁ = 600 K (source), T₂ = 300 K (sink), Q₁ = 2000 J

Step 1: Calculate efficiency
  η = 1 − T₂/T₁ = 1 − 300/600 = 1 − 0.5 = 0.5 = 50%

Step 2: Calculate work done per cycle
  W = η × Q₁ = 0.5 × 2000 = 1000 J

Step 3: Calculate heat rejected
  Q₂ = Q₁ − W = 2000 − 1000 = 1000 J

  Or: Q₂ = Q₁ × (T₂/T₁) = 2000 × (300/600) = 1000 J ✓

Answer: η = 50%, W = 1000 J, Q₂ = 1000 J

3. Derive the expression for work done in an isothermal process and an adiabatic process, and compare them.

Isothermal process (T = constant):

For an ideal gas, PV = nRT. Since T is constant, P = nRT/V.

W = ∫P dV (from V₁ to V₂)
  = ∫(nRT/V) dV
  = nRT ∫(1/V) dV
  = nRT [ln V] from V₁ to V₂
  = nRT ln(V₂/V₁)

Using PV = nRT, this can also be written as:
  W = P₁V₁ ln(V₂/V₁)

Also: W = 2.303 nRT log₁₀(V₂/V₁)

Adiabatic process (ΔQ = 0):

For an adiabatic process, PVᵞ = constant = P₁V₁ᵞ = P₂V₂ᵞ.

W = ∫P dV (from V₁ to V₂)
  = ∫(P₁V₁ᵞ/Vᵞ) dV

  = P₁V₁ᵞ × [V^(1−γ)/(1−γ)] from V₁ to V₂
  = P₁V₁ᵞ/(1−γ) × [V₂^(1−γ) − V₁^(1−γ)]
  = [P₂V₂ − P₁V₁]/(1−γ)     [using PVᵞ = constant]
  = (P₁V₁ − P₂V₂)/(γ − 1)

Since PV = nRT:
  W = nR(T₁ − T₂)/(γ − 1) = nCᵥ(T₁ − T₂)

Also from first law (ΔQ = 0):
  ΔU = −ΔW = nCᵥ(T₂ − T₁)

Comparison:

Property	Isothermal	Adiabatic
Heat exchange	ΔQ ≠ 0	ΔQ = 0
Temperature	Constant	Changes
PV relation	PV = constant	PVᵞ = constant
Slope on PV diagram	Less steep	Steeper (γ > 1)
Internal energy change	ΔU = 0	ΔU = nCᵥΔT
Work done	nRT ln(V₂/V₁)	(P₁V₁ − P₂V₂)/(γ − 1)

Worked Example: 3 moles of an ideal gas at 400 K expand isothermally from 2 L to 4 L. Calculate the work done. If the same expansion occurs adiabatically (γ = 1.4), find the final temperature and work done.

--- Isothermal expansion ---
Given: n = 3, T = 400 K, V₁ = 2 L = 2 × 10⁻³ m³, V₂ = 4 L = 4 × 10⁻³ m³

W = nRT ln(V₂/V₁)
  = 3 × 8.31 × 400 × ln(4/2)
  = 9972 × ln(2)
  = 9972 × 0.693
  = 6910.6 J ≈ 6.91 kJ

--- Adiabatic expansion ---
Given: T₁ = 400 K, V₁ = 2 L, V₂ = 4 L, γ = 1.4

Step 1: Find final temperature
  T₁V₁^(γ−1) = T₂V₂^(γ−1)
  400 × (2)^(0.4) = T₂ × (4)^(0.4)
  400 × 1.3195 = T₂ × 1.7411
  T₂ = 400 × 1.3195 / 1.7411
  T₂ = 527.8 / 1.7411
  T₂ ≈ 303.1 K

Step 2: Calculate work done
  W = nCᵥ(T₁ − T₂)
  Cᵥ = R/(γ − 1) = 8.31/0.4 = 20.775 J/(mol·K)
  W = 3 × 20.775 × (400 − 303.1)
    = 62.325 × 96.9
    = 6039.3 J ≈ 6.04 kJ

Answer: Isothermal work ≈ 6.91 kJ; Adiabatic: T₂ ≈ 303 K, work ≈ 6.04 kJ

Common Mistakes to Avoid

1. Using °C instead of Kelvin in gas laws — All gas law calculations (ideal gas equation, Carnot efficiency, etc.) require absolute temperature in Kelvin. Always convert: K = °C + 273.15.

2. Mixing up sign conventions in the first law — In the first law ΔQ = ΔU + ΔW, heat absorbed by the system is positive, work done by the system is positive. Students often get the sign of ΔW wrong when the gas is compressed (work is done on the gas, so ΔW is negative).

3. Confusing Cₚ and Cᵥ — Cₚ is always greater than Cᵥ by exactly R (Mayer's relation: Cₚ = Cᵥ + R). Cₚ is used for constant-pressure processes; Cᵥ for constant-volume processes. Using Cₚ where Cᵥ is needed (or vice versa) is a common exam error.

4. Applying wrong PV relation for adiabatic process — For isothermal: PV = constant. For adiabatic: PVᵞ = constant. The adiabatic curve is steeper because γ > 1. Do not substitute PV = nRT into the adiabatic relation incorrectly.

5. Forgetting latent heat in calorimetry problems — When a substance changes state (ice to water, water to steam), heat is absorbed or released at constant temperature using Q = mL. Do not use Q = mcΔT during a phase change.

6. Assuming efficiency can reach 100% — Even a Carnot engine has efficiency less than 100% unless T₂ = 0 K (impossible). No real engine can exceed Carnot efficiency. Always express efficiency as a percentage and check if it is between 0 and 100%.

7. Incorrectly calculating work from area under PV curve — In isobaric process, W = PΔV (rectangle). In isothermal process, W = nRT ln(V₂/V₁) (curve). Do not approximate the area as a simple rectangle for isothermal or adiabatic processes.

8. Confusing heat (Q) and internal energy (U) — Heat is energy transferred due to temperature difference; internal energy is the total energy within the system. They are related but distinct. ΔU = nCᵥΔT for an ideal gas regardless of the process.

Exam Checklist

[ ] I can convert between Celsius, Fahrenheit, and Kelvin temperature scales
[ ] I can calculate linear, area, and volume expansion using α, β, and γ
[ ] I can apply Q = mcΔT and Q = mL in calorimetry problems
[ ] I can state the first law of thermodynamics and apply the correct sign convention
[ ] I can derive and use Mayer's relation Cₚ = Cᵥ + R and know Cᵥ, Cₚ values for mono/diatomic gases
[ ] I can calculate work done in isothermal, adiabatic, isobaric, and isochoric processes
[ ] I can derive Carnot efficiency and solve numerical problems involving T₁, T₂, Q₁, Q₂, and W
[ ] I can state the second law in Kelvin-Planck and Clausius forms and explain entropy
[ ] I can distinguish between isothermal and adiabatic processes on a PV diagram
[ ] I can correctly identify state functions (U, S, H) vs path functions (W, Q)

Frequently Asked Questions

The first law states that the heat supplied to a system equals the change in internal energy plus the work done by the system: ΔQ = ΔU + ΔW. It is essentially the law of conservation of energy applied to thermodynamic systems.