Why this topic matters · 9 min read
Heat and Thermodynamics is one of the most consistently tested Physics topics in NDA GAT Paper. Expect 3 to 5 questions per paper covering laws of thermodynamics, heat transfer, specific heat, calorimetry, and thermal expansion. Questions are mostly concept-based with some numerical problems on calorimetry and gas laws. The first and second laws of thermodynamics, Carnot engine efficiency, and modes of heat transfer are high-priority areas. Getting these right is straightforward with the right framework.
Temperature and Heat: The Basics
Heat is energy in transit due to a temperature difference. Temperature is a measure of the average kinetic energy of molecules. They are NOT the same thing. Heat is measured in Joules or Calories. Temperature is measured in Celsius, Kelvin, or Fahrenheit. The SI unit of heat is Joule. One Calorie equals 4.18 Joules — this is the mechanical equivalent of heat given by Joule's experiment.
- Heat flows from higher temperature to lower temperature always.
- 0 Kelvin is absolute zero — no molecular motion below this.
- Kelvin = Celsius + 273 (use 273.15 for precision).
- Fahrenheit = (9/5) x Celsius + 32.
- Triple point of water is 273.16 K — used as a fixed reference in thermometry.
Key formulas
Celsius to Kelvin
T(K) = T(C) + 273
When: Converting temperature for gas law or thermodynamics problems
Joule's Mechanical Equivalent
1 Cal = 4.18 J
When: Converting calories to joules in calorimetry numericals
Worked example
Convert 27 degrees Celsius to Kelvin: T = 27 + 273 = 300 K. This is a very common substitution in gas law problems.
Thermal Expansion
Solids, liquids, and gases expand when heated. Solids have three types of expansion: linear, area, and volumetric. The coefficient of expansion tells how much a material expands per degree rise in temperature. For liquids, only volumetric expansion matters. Gases expand the most and solids the least for the same temperature rise.
- Linear expansion: change in length = original length x alpha x change in T.
- Area expansion coefficient (beta) = 2 x alpha.
- Volume expansion coefficient (gamma) = 3 x alpha.
- Liquids have apparent and real expansion — apparent expansion is less because container also expands.
- Anomalous expansion of water: water expands on cooling from 4 degrees Celsius to 0 degrees Celsius. Maximum density at 4 degrees Celsius.
Key formulas
Linear Expansion
delta L = L0 x alpha x delta T
When: Finding change in length of a rod or rail with temperature change
Volume Expansion
delta V = V0 x gamma x delta T
When: Finding expansion of liquids or gases
Relation between coefficients
gamma = 3 x alpha, beta = 2 x alpha
When: When only one coefficient is given and others are needed
Worked example
A steel rod of length 2 m has alpha = 1.2 x 10^-5 per degree C. Find expansion when heated by 50 degrees C. delta L = 2 x 1.2 x 10^-5 x 50 = 1.2 x 10^-3 m = 1.2 mm.
Calorimetry and Specific Heat
Calorimetry is the science of measuring heat exchange. When two bodies at different temperatures are mixed, heat lost by hotter body equals heat gained by colder body — this is the principle of calorimetry. Specific heat capacity is the heat needed to raise the temperature of 1 kg of a substance by 1 degree Celsius. Latent heat is heat absorbed or released during a change of state without any temperature change.
- Heat gained or lost: Q = m x c x delta T.
- Specific heat of water: 4200 J per kg per degree C (highest among common substances).
- Latent heat of fusion of ice: 336 kJ per kg (melting/freezing).
- Latent heat of vaporisation of water: 2260 kJ per kg (boiling/condensing).
- During change of state, temperature stays constant even though heat is being added or removed.
- Principle of calorimetry: Heat lost = Heat gained (no heat loss to surroundings assumed).
Key formulas
Sensible Heat
Q = m x c x delta T
When: Any heating or cooling problem without phase change
Latent Heat
Q = m x L
When: Melting ice, boiling water, any phase change problem
Calorimetry Principle
m1 x c1 x (T1 - Tf) = m2 x c2 x (Tf - T2)
When: Finding final temperature when two substances are mixed
Worked example
100 g of water at 80 degrees C is mixed with 200 g of water at 20 degrees C. Find final temp: 0.1 x 4200 x (80 - Tf) = 0.2 x 4200 x (Tf - 20). Solving: 80 - Tf = 2(Tf - 20), 80 - Tf = 2Tf - 40, 120 = 3Tf, Tf = 40 degrees C.
Laws of Thermodynamics
Thermodynamics deals with the relationship between heat and work. There are four laws but NDA focuses mainly on the Zeroth, First, and Second Laws. The Zeroth Law defines thermal equilibrium. The First Law is conservation of energy applied to heat and work. The Second Law tells us about the direction of heat flow and efficiency limits of heat engines.
- Zeroth Law: If A is in thermal equilibrium with C, and B is in thermal equilibrium with C, then A and B are in thermal equilibrium with each other. This defines temperature.
- First Law: delta Q = delta U + delta W. Heat added to system equals increase in internal energy plus work done by system.
- Second Law: Heat cannot spontaneously flow from cold to hot. No engine can be 100 percent efficient.
- Isothermal process: Temperature constant, delta U = 0, so Q = W.
- Adiabatic process: No heat exchange, Q = 0, so delta U = -W.
- Isobaric: Pressure constant. Isochoric (Isovolumetric): Volume constant, W = 0, so Q = delta U.
Key formulas
First Law of Thermodynamics
delta Q = delta U + delta W
When: Any thermodynamic process to find unknown among heat, internal energy, work
Work done by gas
W = P x delta V (at constant pressure)
When: Isobaric process only
Carnot Engine and Efficiency
A Carnot engine is an ideal heat engine working between two temperatures. It has the maximum possible efficiency for any engine working between the same two temperatures. No real engine can match Carnot efficiency. NDA regularly asks numerical questions on this formula. Remember: temperatures MUST be in Kelvin.
- Carnot efficiency depends only on source and sink temperatures.
- Higher the temperature difference, higher the efficiency.
- Efficiency is always less than 1 (100 percent) for any real engine.
- COP of refrigerator = T2 divided by (T1 - T2), where T2 is cold reservoir.
- Carnot engine is reversible and operates on Carnot cycle: two isothermals and two adiabatics.
Key formulas
Carnot Efficiency
eta = 1 - (T2 / T1) = (T1 - T2) / T1
When: Finding efficiency of ideal engine; T1 = source temp, T2 = sink temp, both in Kelvin
Worked example
A Carnot engine works between 500 K and 300 K. Efficiency = 1 - (300/500) = 1 - 0.6 = 0.4 = 40 percent.
Modes of Heat Transfer
Heat transfers in three ways: Conduction, Convection, and Radiation. Conduction is transfer through a solid without movement of particles. Convection is transfer through fluids (liquids and gases) by actual movement of particles. Radiation is transfer through electromagnetic waves and does NOT need a medium — this is how the Sun heats the Earth.
- Conduction: Solids only (metals are best conductors). Fourier's Law governs it.
- Convection: Fluids only (liquids and gases). Sea breeze and land breeze are natural convection examples.
- Radiation: No medium needed. All bodies above absolute zero radiate heat.
- Stefan-Boltzmann Law: Power radiated is proportional to T raised to the power 4.
- Black body absorbs all radiation (absorptivity = 1). Good absorbers are good emitters.
- Newton's Law of Cooling: Rate of heat loss is proportional to the excess temperature over surroundings.
Key formulas
Stefan-Boltzmann Law
P = sigma x A x T^4 (for black body)
When: Radiation problems comparing power emitted at different temperatures
Newton's Law of Cooling
dQ/dt = -k(T - T0)
When: Rate of cooling problems where excess temperature is given
⚠ Common mistakes to avoid
- Using Celsius instead of Kelvin in Carnot efficiency formula — always convert to Kelvin first or the answer will be completely wrong.
- Confusing heat and internal energy in First Law — delta Q is heat added TO the system, delta W is work done BY the system, not on it.
- Forgetting that during a change of state (melting or boiling) temperature does NOT change — many students apply Q = mcT here instead of Q = mL.
- Mixing up real and apparent expansion of liquids — apparent expansion is what we observe in a container, real expansion is always larger.
- Assuming higher specific heat means a substance heats up faster — it is exactly the opposite. Higher specific heat means slower heating and slower cooling.
🧠 Memory aids
- Mnemonic for Thermodynamic Laws — You Cannot Break Even, And You Must Play: Zeroth = You (define temp), First = Cannot (energy conserved, cannot create), Second = Break Even (cannot be 100 percent efficient), Third = Must Play (cannot reach absolute zero).
- For expansion coefficients: Remember GAB = Gamma is 3 Alpha, Beta is 2 Alpha. Think of it as going up in steps: L A 2A 3A.
- Modes of heat transfer — CCR: Conduction needs Contact, Convection needs Currents (fluid flow), Radiation needs nothing (Remote).
- Calorimetry check: LOST = GAINED. Hot body loses, cold body gains. Set them equal and solve for unknown.
🎯 NDA exam tips
- NDA asks 1 to 2 direct formula-based numericals on Carnot efficiency almost every year — just memorise the formula and practice converting Celsius to Kelvin before substituting.
- Conceptual questions on the Second Law such as which process is impossible or which engine is most efficient are very common — know that Carnot is always the most efficient between same two temperatures.
- Anomalous expansion of water and its significance for aquatic life in winter is a favourite one-liner question in GAT — do not skip it.
- Questions on modes of heat transfer are mostly identification type — which mode operates in a vacuum (radiation), why metals feel colder than wood at same temperature (conduction), etc.
- For calorimetry numericals, the mix of ice and water type questions come frequently — remember to account for latent heat when ice melts first before temperature rises, which is a two-step calculation many aspirants miss.
Q1 · medium · AI-verified
A gas expands from 2 L to 8 L at constant pressure of 3 atm. The work done by the gas is:
- 6 J
- 18 J
- 1824 J
- 608 J
Q2 · medium · AI-verified
An ideal gas undergoes an isothermal expansion from volume V to 3V at temperature 300 K. If the initial pressure was 6 atm, what is the final pressure of the gas?
- 18 atm
- 1 atm
- 3 atm
- 2 atm
Q3 · medium · AI-verified
The efficiency of a Carnot engine operating between temperatures 400 K and 300 K is:
- 25%
- 75%
- 33%
- 66%
Q4 · medium · AI-verified
A system absorbs 300 J of heat and does 200 J of work. The change in internal energy is:
- 100 J
- 500 J
- -100 J
- 0 J
Q5 · medium · AI-verified
For a reversible heat engine operating between two reservoirs, the entropy change of the universe is:
- positive
- negative
- zero
- infinite