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First Law of Thermodynamics

Calculate ΔU = Q − W from heat transfer and work

Parameters

Jⓘ
Jⓘ
Show Trail

Controls

xⓘ

Calculated Values

ΔU (Internal Energy Change):
300.00;J300.00;J
Heat to Work Ratio:
2.50;2.50;

Examples

Heating with expansion

Q = 500 J, W = 200 J.

  • ΔU: 300.00300.00

Compression

Q = 0, W = −100 J (work on system).

  • ΔU: 100.00100.00

Visualization

First Law of Thermodynamics — Conservation of Energy

The first law of thermodynamics is the law of energy conservation applied to heat and work. For a closed system: ΔU = Q − W, where ΔU is the change in internal energy, Q is heat added to the system, and W is work done by the system on the surroundings. This is the foundation of all thermal physics and engineering thermodynamics.

Sign convention (physics/engineering): Q > 0 when heat flows into the system; W > 0 when the system expands and pushes the piston outward (does work on surroundings). If work is done on the system (compression), W < 0 and internal energy rises. Chemistry texts sometimes use W_on (work on system) with ΔU = Q + W_on — always check which convention is used.

Internal energy U is a state function: it depends only on the current state (T, P, V for a simple compressible system), not on how that state was reached. Heat Q and work W are path functions — the same ΔU can result from heating alone, compression alone, or a combination.

For an ideal gas with fixed composition, Joule's law gives U = U(T) only, so ΔU = nC_vΔT. Mayer's relation C_p − C_v = R links heat capacities. Isochoric (constant V): W = 0 ⇒ Q = ΔU. Isobaric: W = PΔV and Q = ΔH = ΔU + PΔV. Adiabatic: Q = 0 ⇒ ΔU = −W.

Cyclic processes return the system to its initial state, so ΔU_cycle = 0 and Q_net = W_net. The area enclosed on a P–V diagram equals net work per cycle. Isolated systems (adiabatic rigid walls, no work) have Q = W = 0 and ΔU = 0.

The first law forbids perpetual motion machines of the first kind (devices that produce net work without energy input). It does not forbid converting heat to work (engines) or work to heat (friction) — only that energy is accounted for. Enthalpy H = U + PV is often more useful than U for constant-pressure chemical reactions and open-flow systems.

Key Concepts

  • ΔU = Q − W (system-centric: heat in, work out)
  • U is a state function; Q and W depend on path
  • Ideal gas: ΔU = nC_vΔT; C_p − C_v = R
  • Cyclic process: ΔU = 0 ⇒ Q_net = W_net
  • Isolated system: Q = W = 0, ΔU = 0
  • Enthalpy H = U + PV for constant-P processes

Real-World Applications

  • Internal combustion and steam engine energy balances
  • Refrigerator and heat pump COP analysis (first law + second law)
  • Calorimetry and bomb calorimeter reaction energy
  • Compressed air storage (work ↔ internal energy)
  • Class 11–12 CBSE/NCERT thermodynamics problems
  • Atmospheric parcel expansion (approximate energy partitioning)

Explore Further

More thermodynamics tools

Physics Equations

First Law:
ΔU=Q−W\Delta U = Q - W
Energy Balance:
Q=ΔU+WQ = \Delta U + W

Step-by-Step Solution

See how the main results are calculated.

1

Step 1: Sign Convention

Define Q (heat in) and W (work out by system).

Explanation:

Q > 0: heat enters system. W > 0: system does work on surroundings (expansion).

2

Step 2: First Law

Equation:

ΔU=Q−W\Delta U = Q - W

Explanation:

Internal energy change equals heat added minus work done by the system.

3

Step 3: Substitute

Calculation:

ΔU=500−(200)=300.0000 J\Delta U = 500 - (200) = 300.0000 \text{ J}

Result:

ΔU=300.0000JΔU = 300.0000 J

Explanation:

ΔU > 0 means internal energy increased (temperature may rise for ideal gas).

4

Step 4: Interpret

Relate to temperature change for ideal gas.

Explanation:

For ideal gas, ΔU = nC_vΔT. Positive ΔU usually means temperature increase if no phase change.

5

Step 5: Energy Balance Check

Verify conservation.

Calculation:

Q=ΔU+W=300.0000+200=500.0000Q = \Delta U + W = 300.0000 + 200 = 500.0000

Explanation:

Heat in partitions into stored energy and work out.

6

Step 6: Process Type Hint

Classify the process if possible.

Explanation:

W = 0 → isochoric; Q = 0 → adiabatic; Q = W with ΔU = 0 → isothermal (ideal gas).

Frequently Asked Questions (FAQ)

Why Q − W and not Q + W?

Convention: W positive when system expands (loses energy as work). Other texts use ΔU = Q + W_on with opposite W sign — stay consistent.

Can ΔU be negative?

Yes — system cools or loses internal energy when W > Q.

Is heat a state function?

No — heat depends on path. Only ΔU (and U) are state functions.

First law and perpetual motion?

You cannot create energy; first law forbids machines that produce net energy without input.

Relation to enthalpy?

H = U + PV useful for constant-pressure chemistry; first law still underlies energy balance.

Practice MCQs

  1. If Q = 200 J enters and W = 50 J is done by the system, ΔU is:
  2. In an adiabatic process:
  3. For a complete cycle, ΔU equals:
  4. Work done by the system on surroundings is:
  5. Internal energy is a:
  6. Isochoric heating of ideal gas: W = 0 implies:

Related Calculators

These tools connect to the same physics concepts used in this calculator.

Adiabatic ProcessFind P₂ and T₂ for reversible adiabatic ideal gas expansion (Q = 0).Isothermal ProcessCalculate work W = nRT ln(V₂/V₁) and heat for isothermal ideal gas.Boyle's LawCalculate pressure or volume at constant temperature (P₁V₁ = P₂V₂).Charles's LawCalculate volume change with temperature at constant pressure.Error PropagationCombine measurement uncertainties for sums, products, quotients, and powers in experimental formulas.Related Measurement & Uncertainty topic.Maxwell-Boltzmann DistributionCalculate velocity distributions and characteristic velocities for ideal gases using Maxwell-Boltzmann statistics.Related Statistical Physics topic.Projectile MotionCalculate and visualize the path of objects under the influence of gravity.Related Mechanics topic.