Playground
Interactive energy flow diagram: adjust heat input Q and work output W to see internal energy change with animated arrows and bar chart.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Change in internal energyoutput Net change in the system's internal energy | J | M·L²·T⁻² | -1000 – 1000 | |
| Heat added Heat energy transferred into the system | J | M·L²·T⁻² | -1000 – 1000 | |
| Work done by system Work done by the system on its surroundings | J | M·L²·T⁻² | -1000 – 1000 |
Deep dive
Derivation
From conservation of energy: the total energy of an isolated system is constant. For a closed system exchanging heat Q and doing work W on surroundings: ΔU = Q − W. This is an axiom (not derived from deeper principles) but validated by Joule's paddle-wheel experiment showing 1 cal = 4.186 J.
Experimental verification
Joule's paddle-wheel experiment (1845) measured the mechanical equivalent of heat: 4.186 J/cal. Modern calorimetry confirms energy conservation to parts per billion in nuclear reactions (E = mc²) and chemical reactions.
Common misconceptions
- Heat and work are not properties of a system — they are processes (path-dependent), while internal energy U is a state function
- Sign convention varies: physics uses W = work BY system (ΔU = Q − W), engineering often uses W = work ON system (ΔU = Q + W)
- Temperature change does not always mean heat was added — adiabatic compression raises temperature with Q = 0
Real-world applications
- Engine cycle analysis (Otto, Diesel, Rankine cycles)
- Refrigerator and heat pump efficiency calculations
- Metabolic energy accounting in biology
- Chemical reactor energy balance in industrial processes
Worked examples
Bicycle pump heating
Given:
- Q:
- 500
- W:
- 200
Find: Delta_U
Solution
ΔU = Q − W = 500 − 200 = 300 J
Adiabatic compression
Given:
- Q:
- 0
- W:
- -150
Find: Delta_U
Solution
ΔU = Q − W = 0 − (−150) = 150 J
Scenarios
What if…
- scenario:
- What if Q = W (isothermal process for an ideal gas)?
- answer:
- Then ΔU = 0. All heat input becomes work output. The gas expands at constant temperature — its internal energy doesn't change because U depends only on T for an ideal gas.
- scenario:
- What if the system is perfectly insulated (Q = 0)?
- answer:
- ΔU = −W. Any work done by the gas comes entirely from its internal energy, cooling it down. This is how adiabatic cooling works in expanding gases (e.g., canned air getting cold when sprayed).
Limiting cases
- condition:
- Q = 0 (adiabatic)
- result:
- ΔU = −W
- explanation:
- No heat exchange: work done by gas decreases internal energy.
- condition:
- W = 0 (isochoric)
- result:
- ΔU = Q
- explanation:
- Constant volume: all heat goes into internal energy.
- condition:
- ΔU = 0 (isothermal, ideal gas)
- result:
- Q = W
- explanation:
- Temperature unchanged: all heat input becomes work output.
Context
Rudolf Clausius · 1850
Clausius formalized the relationship between heat and work, building on Joule's experiments showing mechanical equivalence of heat and Mayer's earlier theoretical work.
Hook
If you pump a bicycle tire, why does the pump get hot?
A gas absorbs 500 J of heat and does 200 J of work on a piston. Find the change in internal energy.
Dimensions: [ΔU] = [Q] − [W] → M·L²·T⁻² = M·L²·T⁻² − M·L²·T⁻² ✓ (all terms are energy in joules)
Validity: Universal law with no known exceptions. Applies to all thermodynamic systems. Sign conventions vary by textbook (physics: W = work by system; engineering: W = work on system).