First Law of Thermodynamics

First Law of Thermodynamics

Energy Conservation — Formula, Derivation, Worked Examples & Exam Insights

Last Updated: March 2026

📌 Key Takeaways

  • Core principle: Energy can neither be created nor destroyed — only transformed from one form to another.
  • Closed system formula: Q = ΔU + W (heat added = change in internal energy + work done by system).
  • Open system formula (SFEE): Q − W = Δh + ΔKE + ΔPE, accounting for mass flow across boundaries.
  • Heat (Q) and Work (W) are path functions — their values depend on how the process occurs. Internal energy (U) is a state function.
  • The First Law tells us how much energy is involved but not which direction a process will go — that requires the Second Law.
  • Sign convention (engineering): Q is positive when heat is added to the system; W is positive when work is done by the system.

1. Statement of the First Law

The First Law of Thermodynamics is the principle of conservation of energy applied to thermodynamic systems. It states:

Energy can neither be created nor destroyed. It can only be transferred or converted from one form to another. The total energy of an isolated system remains constant.

This means every joule of heat you add to a system must show up as either an increase in the system’s stored energy (internal energy) or as work done by the system on its surroundings — or some combination of both. No energy disappears, and no energy appears from nothing.

The First Law is sometimes called the law of conservation of energy in a thermodynamic context. It was established through the pioneering work of James Prescott Joule in the 1840s, who demonstrated through careful experiments that mechanical work and heat are interconvertible — a fixed amount of work always produces the same amount of heat.

2. First Law for Closed Systems

A closed system is one where no mass crosses the system boundary — only energy (as heat or work) can enter or leave. For a closed system undergoing a process from state 1 to state 2:

First Law — Closed System

Q = ΔU + W

Or equivalently: Q − W = ΔU

Where:

  • Q = net heat added to the system (J or kJ)
  • ΔU = U&sub2; − U&sub1; = change in internal energy (J or kJ)
  • W = net work done by the system (J or kJ)

This equation is the mathematical form of the First Law for closed systems. Every thermodynamic problem involving a closed system starts here.

For a cyclic process (system returns to its initial state), ΔU = 0, so:

First Law — Cyclic Process

Qnet = Wnet

In a complete cycle, the net heat added equals the net work done. This is the basis of all heat engines.

3. Sign Convention — Getting It Right

Sign convention errors are the most common source of mistakes in thermodynamics problems. Two conventions exist — make sure you know which one your textbook and exam use:

QuantityEngineering Convention (Most Indian Textbooks)Physics / IUPAC Convention
Heat (Q)Positive when added to systemPositive when added to system
Work (W)Positive when done by the systemPositive when done on the system
First Law formQ = ΔU + WΔU = Q + W

This guide uses the engineering convention (Q = ΔU + W) because it is the standard in most Indian engineering textbooks, GATE, and ESE papers. Work done by the system is positive, work done on the system is negative.

Exam tip: Before solving any problem, write down your sign convention explicitly. Many students lose marks not because they do not understand the physics, but because they mix up signs midway through a calculation.

4. Internal Energy — What the System Stores

Internal energy (U) is the total energy stored within a system at the microscopic level. It includes:

  • Kinetic energy of molecules: Translational, rotational, and vibrational motion of atoms and molecules.
  • Potential energy of molecular interactions: Intermolecular forces, chemical bonds.

Internal energy does not include the kinetic or potential energy of the system as a whole (that is accounted for separately in open system analysis).

Key properties of internal energy:

PropertyExplanation
State functionU depends only on the current state (P, V, T) — not on how the system reached that state.
Extensive propertyU depends on the amount of substance. Specific internal energy (u = U/m) is intensive.
For ideal gasesU depends only on temperature: ΔU = nCvΔT. This greatly simplifies calculations.

Internal Energy Change — Ideal Gas

ΔU = nCvΔT = nCv(T&sub2; − T&sub1;)

Where: n = number of moles, Cv = molar specific heat at constant volume (J/mol·K), ΔT = temperature change (K)

5. First Law Applied to Common Processes

Different thermodynamic processes impose different constraints. Here is how the First Law simplifies for each:

ProcessConstraintFirst Law BecomesKey Insight
Isochoric (constant volume)V = constant → W = 0Q = ΔUAll heat goes into changing internal energy. No work done.
Isobaric (constant pressure)P = constant → W = PΔVQ = ΔU + PΔV = ΔHHeat added equals enthalpy change. This is why enthalpy matters for constant-pressure processes.
Isothermal (constant temperature)T = constant → ΔU = 0 (ideal gas)Q = WAll heat added is converted to work. Internal energy unchanged for ideal gases.
Adiabatic (no heat transfer)Q = 0W = −ΔUWork comes at the expense of internal energy. System cools during expansion, heats during compression.
Free expansion (into vacuum)W = 0, Q = 0ΔU = 0No work against vacuum, no heat transfer. Temperature unchanged for ideal gases.

6. First Law for Open Systems — Steady Flow Energy Equation (SFEE)

An open system (control volume) allows mass to flow in and out across its boundary. Examples include turbines, compressors, nozzles, heat exchangers, and pumps. The First Law for a steady-flow open system is:

Steady Flow Energy Equation (SFEE)

Q − W = ṁ [(h&sub2; − h&sub1;) + (V&sub2;² − V&sub1;²)/2 + g(z&sub2; − z&sub1;)]

Where:

  • Q = rate of heat transfer (W or kW)
  • W = rate of shaft work (W or kW)
  • ṁ = mass flow rate (kg/s)
  • h = specific enthalpy (kJ/kg)
  • V = velocity (m/s)
  • g = gravitational acceleration (9.81 m/s²)
  • z = elevation (m)

For most engineering devices, specific terms can be neglected based on the situation:

DeviceSFEE SimplificationNeglected Terms
Nozzle / Diffuserh&sub1; + V&sub1;²/2 = h&sub2; + V&sub2;²/2Q ≈ 0, W = 0, ΔPE ≈ 0
Turbine / CompressorQ − W = ṁ(h&sub2; − h&sub1;)ΔKE ≈ 0, ΔPE ≈ 0
Heat ExchangerQ = ṁ(h&sub2; − h&sub1;)W = 0, ΔKE ≈ 0, ΔPE ≈ 0
Throttling Valveh&sub1; = h&sub2; (isenthalpic)Q ≈ 0, W = 0, ΔKE ≈ 0, ΔPE ≈ 0

7. Worked Numerical Examples

Example 1: Closed System — Gas Expansion

Problem: A gas in a piston-cylinder assembly receives 500 kJ of heat and does 200 kJ of work on the surroundings. Find the change in internal energy.

Solution

Given: Q = +500 kJ (heat added), W = +200 kJ (work done by system)

Using First Law: Q = ΔU + W

500 = ΔU + 200

ΔU = 300 kJ

The internal energy of the gas increases by 300 kJ. Of the 500 kJ of heat added, 200 kJ was used to do work and 300 kJ was stored as internal energy.

Example 2: Adiabatic Compression

Problem: 2 moles of an ideal gas (Cv = 20.8 J/mol·K) are compressed adiabatically. 800 J of work is done on the gas. Find the temperature change.

Solution

Given: Q = 0 (adiabatic), W = −800 J (work done ON the system, so negative in engineering convention), n = 2 mol

First Law: Q = ΔU + W → 0 = ΔU + (−800) → ΔU = +800 J

For ideal gas: ΔU = nCvΔT

800 = 2 × 20.8 × ΔT

ΔT = 19.23 K

The temperature increases by about 19.2 K. This makes physical sense — compressing a gas adiabatically raises its temperature.

Example 3: Open System — Steam Turbine

Problem: Steam enters a turbine at h&sub1; = 3,200 kJ/kg and exits at h&sub2; = 2,600 kJ/kg. The mass flow rate is 5 kg/s. Heat loss from the turbine casing is 50 kW. Find the power output. Neglect KE and PE changes.

Solution

Using simplified SFEE: Q − W = ṁ(h&sub2; − h&sub1;)

Q = −50 kW (heat lost from system, so negative)

−50 − W = 5 × (2600 − 3200)

−50 − W = 5 × (−600) = −3000

−W = −3000 + 50 = −2950

W = 2,950 kW = 2.95 MW

The turbine produces 2.95 MW of power. The 50 kW heat loss slightly reduces the output from what would be 3 MW in an adiabatic case.

8. What the First Law Cannot Tell You

The First Law is powerful but has a critical limitation: it says nothing about the direction of a process.

Consider these two scenarios — both satisfy the First Law:

  • A hot cup of tea cools down in a room, transferring heat to the air. (Q = ΔU + W is satisfied.)
  • A cup of tea spontaneously heats up by absorbing heat from the cooler room air. (Q = ΔU + W is also satisfied — energy is conserved.)

The First Law allows both. But in reality, only the first scenario happens. Hot objects cool down; they never spontaneously heat up by extracting energy from cooler surroundings. The reason is the Second Law of Thermodynamics, which introduces entropy and tells us which processes are possible and which are not.

Similarly, the First Law cannot tell you the maximum possible efficiency of a heat engine — only the Second Law (via the Carnot cycle) can set that upper bound.

In summary: the First Law tells you how much, the Second Law tells you which way.

9. Common Mistakes Students Make

  • Mixing up sign conventions: The most frequent error. Some textbooks use Q = ΔU + W (engineering convention), others use ΔU = Q + W (physics convention, where work done ON the system is positive). Mixing them gives wrong signs. Always write your convention at the start of a problem.
  • Treating heat or work as state functions: Q and W depend on the path taken. Only ΔU is path-independent. You cannot say “the heat of a system is 500 kJ” — heat is a transfer quantity, not a property of the system.
  • Forgetting that ΔU = 0 for isothermal processes applies only to ideal gases: For real gases, internal energy can change even at constant temperature because of intermolecular forces. The simplification ΔU = 0 at constant T is strictly valid only for ideal gas behaviour.
  • Neglecting terms in SFEE without justification: Students sometimes drop kinetic energy, potential energy, or heat transfer terms without stating why. In exams, always state which terms you are neglecting and why (e.g., “KE change is negligible because inlet and outlet velocities are similar”).
  • Confusing work done BY the system with work done ON the system: If a gas expands and pushes a piston, the system does work on the surroundings (W positive in engineering convention). If a piston compresses the gas, work is done on the system (W negative). Getting this backwards inverts the answer.

10. Frequently Asked Questions

What does the First Law of Thermodynamics state?

The First Law states that energy cannot be created or destroyed — it can only be transferred between a system and its surroundings as heat or work, or converted from one form to another. Mathematically, for a closed system: Q = ΔU + W. The total energy of an isolated system is always constant.

What is the difference between the First Law for closed and open systems?

For a closed system (no mass flow), the First Law is Q = ΔU + W. For an open system (mass enters and exits), the Steady Flow Energy Equation (SFEE) applies: Q − W = ṁ[(h&sub2; − h&sub1;) + ΔKE + ΔPE]. The open system form uses enthalpy (h) instead of internal energy (U) because enthalpy accounts for the flow work required to push mass across the system boundary.

Can energy be created or destroyed according to the First Law?

No. This is the fundamental assertion of the First Law. Energy can change form — chemical to thermal, thermal to mechanical, electrical to mechanical — but the total quantity is always conserved. No experiment in the history of science has ever observed a violation of energy conservation in a macroscopic system.

Why can’t the First Law alone determine the direction of a process?

The First Law only enforces an energy balance — it ensures the books add up. It does not tell you which direction the energy will naturally flow. A hot body losing heat to cold surroundings satisfies the First Law, and so does a cold body losing heat to hot surroundings (energy is conserved either way). Only the Second Law introduces directionality through the concept of entropy increase.