Latest Update: SSC CGL 2025 Notification Out!
Mechanical Engineering (SSC JE) Notes

Thermodynamics Notes for SSC JE Mechanical Engineering

Rahul Kumar

SSC Exam Expert & Content Editor

Updated 21 Oct 2026

PDF Notes: Thermodynamics Notes for SSC JE Mechanical Engineering

Premium PDF · one-time payment of ₹5 · instant access

Thermodynamics studies heat, work and energy and how they change from one form to another. It is the most important part of the Thermal Engineering section in the SSC JE Mechanical paper, and gives both theory and numerical questions. These thermodynamics notes for SSC JE explain the laws, the processes, the gas formulas and the cycles in simple language with solved examples and practice questions.

Basic Terms

TermMeaning
SystemThe part of the universe chosen for study
Closed systemEnergy can cross the boundary, but mass cannot
Open systemBoth mass and energy can cross the boundary
Isolated systemNeither mass nor energy can cross the boundary
Intensive propertyDoes not depend on mass: pressure, temperature, density
Extensive propertyDepends on mass: volume, energy, entropy

Laws of Thermodynamics

LawStatement
Zeroth lawIf two bodies are in thermal equilibrium with a third body, they are in equilibrium with each other; it gives the idea of temperature
First lawEnergy can neither be created nor destroyed; ΔU = Q − W
Second law (Kelvin-Planck)No heat engine can convert all the heat into work; no engine has 100 per cent efficiency
Second law (Clausius)Heat cannot flow from a cold body to a hot body by itself without outside work
Third lawEntropy of a perfect crystal is zero at absolute zero temperature

Thermodynamic Processes

ProcessConditionWork done
IsochoricVolume is constantW = 0
IsobaricPressure is constantW = P (V2 − V1)
IsothermalTemperature is constant; PV = constantW = P1V1 ln (V2 / V1)
AdiabaticNo heat transfer (Q = 0); PVγ = constantW = (P1V1 − P2V2) / (γ − 1)
PolytropicPVn = constantW = (P1V1 − P2V2) / (n − 1)

Ideal Gas Relations

RelationFormula
Ideal gas equationPV = mRT
Specific heatsCp − Cv = R
Ratio of specific heatsγ = Cp / Cv (1.4 for air)
Cv and CpCv = R / (γ − 1) and Cp = γR / (γ − 1)
Gas constant of airR = 287 J/kg K
EnthalpyH = U + PV

Heat Engine, Refrigerator and Heat Pump

DeviceFormula
Efficiency of heat engineη = W / Q1 = 1 − Q2 / Q1
Carnot efficiencyη = 1 − T2 / T1 (kelvin)
COP of refrigeratorQ2 / W = T2 / (T1 − T2) for a Carnot refrigerator
COP of heat pumpQ1 / W = T1 / (T1 − T2)
RelationCOP of heat pump = COP of refrigerator + 1

Example 1. A Carnot engine works between 600 K and 300 K. Find its efficiency.
η = 1 − 300 / 600 = 0.5 or 50%. If it takes 1000 J of heat, it gives 500 J of work.

Example 2. A Carnot refrigerator works between 270 K and 300 K. Find its COP.
COP = 270 / (300 − 270) = 9.

Air Standard Cycles

CycleUsed inImportant point
Carnot cycleIdeal engineTwo isothermal and two adiabatic processes; the maximum possible efficiency
Otto cyclePetrol (spark ignition) engineη = 1 − 1 / rγ − 1, where r is the compression ratio
Diesel cycleDiesel (compression ignition) engineHeat is added at constant pressure
Dual cycleModern high-speed diesel enginesHeat is added partly at constant volume and partly at constant pressure
Brayton cycleGas turbineTwo isobaric and two adiabatic processes
  • For the same compression ratio and the same heat input, the Otto cycle is more efficient than the Diesel cycle.
  • For the same maximum pressure and heat input, the Diesel cycle is more efficient.

Entropy and Steam

  • Entropy is a measure of the disorder or the unavailable energy of a system; for a reversible process, dS = dQ / T.
  • In an isolated system, entropy never decreases (the principle of increase of entropy).
  • The critical point of water is at 374.15 degrees Celsius and 221.2 bar, where the liquid and the vapour become the same.
  • The triple point of water is at 273.16 K.
  • Dryness fraction (x) = mass of dry steam / total mass of the steam-water mixture.

Quick Tips

  • Always convert the temperature into kelvin in the formulas of efficiency and COP.
  • Remember that the heat of the exhaust is the reason why no engine is 100 per cent efficient.
  • Learn the work done in each process in a table.
  • Remember that the Carnot cycle has the highest efficiency between two given temperatures.

Quick Revision Points

  • First law: ΔU = Q − W.
  • Carnot efficiency = 1 − T2 / T1.
  • Cp − Cv = R and γ = Cp / Cv.
  • Isochoric process has zero work, and adiabatic process has zero heat transfer.
  • Otto cycle efficiency = 1 − 1 / rγ − 1.

Practice Questions With Answers

  1. What is the value of work done in an isochoric process?
    Answer: Zero
  2. Which law gives the idea of temperature?
    Answer: Zeroth law
  3. What is the relation between Cp, Cv and R?
    Answer: Cp − Cv = R
  4. A Carnot engine works between 800 K and 400 K. Find the efficiency.
    Answer: 50%
  5. What is the COP of a Carnot refrigerator working between 250 K and 300 K?
    Answer: 5 (250 / 50)
  6. Which cycle is used in a petrol engine?
    Answer: Otto cycle
  7. Which cycle has heat addition at constant pressure?
    Answer: Diesel cycle
  8. What is the value of γ for air?
    Answer: 1.4
  9. Which process has no heat transfer?
    Answer: Adiabatic process
  10. What is the critical temperature of water?
    Answer: 374.15 degrees Celsius

How to Practise This Chapter

Solve three numericals daily on efficiency, COP and processes, and keep a formula sheet. Then solve questions from our practice papers, the free SSC quizzes and the previous year papers page. Read the next chapter, Strength of Materials, and see all topics on the SSC JE Mechanical Engineering notes page. For basic physics, also read the General Science notes.

Frequently Asked Questions

What is thermodynamics?

Thermodynamics is the branch of science that studies heat, work and energy and the way they change from one form to another.

What is the first law of thermodynamics?

The first law says that energy cannot be created or destroyed. For a system, the change in internal energy equals heat added minus work done by the system.

What is the second law of thermodynamics?

The second law says that no heat engine can convert all the heat into work, and heat cannot flow from cold to hot by itself.

What is the zeroth law of thermodynamics?

If two bodies are in thermal equilibrium with a third body, they are in thermal equilibrium with each other.

What is an adiabatic process?

An adiabatic process is a process in which there is no heat transfer between the system and the surroundings.

What is the efficiency of the Carnot cycle?

The efficiency of the Carnot cycle is 1 minus the ratio of the sink temperature to the source temperature in kelvin.

What is COP?

COP is the coefficient of performance of a refrigerator or heat pump. It is the useful heat effect divided by the work input.

Which cycle is more efficient, Otto or Diesel?

For the same compression ratio, the Otto cycle is more efficient. For the same maximum pressure, the Diesel cycle is more efficient.

What is entropy?

Entropy is a property that measures the disorder or the unavailable energy of a system.

What is the critical point of water?

It is the point at 374.15 degrees Celsius and 221.2 bar, where the liquid and the vapour phases become the same.

Is thermodynamics important for SSC JE Mechanical?

Yes. It is one of the most important parts of the Thermal Engineering section and gives both theory and numerical questions.

How many questions come from thermodynamics in SSC JE?

Usually four to eight questions come from thermodynamics and thermal engineering, but the number changes from year to year.

About the Author

Rahul Kumar

SSC Exam Expert & Content Editor

Rahul is a senior SSC exam strategist with 8+ years of experience helping aspirants crack CGL, CHSL and MTS exams. He writes in-depth notification breakdowns, exam pattern guides and preparation strategies.