Thermodynamic processes and heat engines
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All 12 Flashcards — Thermodynamic processes and heat engines
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Question
State the first law of thermodynamics.
Answer
$\Delta U = Q - W$, where **W** is the work done **by** the gas. Internal energy U depends only on temperature.
Question
Isothermal process — what is constant, and the consequence?
Answer
**T** is constant, so $\Delta U = 0$ and therefore $Q = W$.
Question
Isobaric process — what is constant, and the work?
Answer
**P** is constant; the work done by the gas is $W = P\,\Delta V$.
Question
Isovolumetric process — what is constant, and the consequence?
Answer
**V** is constant, so $W = 0$ and therefore $Q = \Delta U$.
Question
Adiabatic process — what is zero, and the consequence?
Answer
**Q = 0** (no heat flows), so $\Delta U = -W$.
Question
On a p–V diagram, what is the work done by the gas?
Answer
The **area under the curve** between the start and end volumes.
Question
What does a heat engine do each cycle?
Answer
Takes in **Q_in** from the hot reservoir, does useful **work W**, and rejects **Q_out** to the cold reservoir. $W = Q_{in} - Q_{out}$.
Question
Give the efficiency formula for a heat engine.
Answer
$\eta = \dfrac{\text{useful work}}{\text{energy input}} = 1 - \dfrac{Q_{out}}{Q_{in}}$.
Question
Give the Carnot (maximum) efficiency formula.
Answer
$\eta_{Carnot} = 1 - \dfrac{T_{cold}}{T_{hot}}$, with both temperatures in **kelvin**.
Question
Why is a real engine's efficiency below the Carnot value?
Answer
Friction, turbulence and unwanted heat loss waste energy, so the real efficiency is always **lower** than the Carnot ceiling.
Question
Worked example — efficiency from Q_in = 800 J, Q_out = 600 J?
Answer
$\eta = 1 - \dfrac{600}{800} = 0.25$, i.e. **25%**.
Question
Worked example — Carnot efficiency between 500 K and 300 K?
Answer
$\eta_{Carnot} = 1 - \dfrac{300}{500} = 0.40$, i.e. **40%**.
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Topic 2.4 hub
Thermodynamics (HL)
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