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Waste heat and the Carnot limit

Use Carnot limits to quantify unavoidable waste heat and understand the thermal bottleneck in real engines.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Thermal Bottleneck. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

The thermal bottleneck is the unavoidable efficiency limit when useful work is produced through heat flow between finite temperatures.

At a glance

  • Prerequisites: The Carnot engine
  • Outcomes: estimate best-case efficiency and unavoidable waste heat for a given Tₕ, Tc
  • Key result: eₘₐₓ = 1-Tc/Tₕ, (Q_(c, min))/Qₕ = Tc/Tₕ
Common traps
  • Treating a first-law limit (W ≤ Qₕ) as if it were the real limit; the second law adds the tighter Carnot bound. - Using Tc/Tₕ with temperatures in °C instead of K. - Thinking “waste heat” is only bad design; some rejection is fundamentally unavoidable when Tc > 0.

Setup (model, assumptions, sign conventions)

Model a heat engine between two reservoirs:

  • hot reservoir at Tₕ
  • cold reservoir at Tc
  • heat input Qₕ
  • work output Wₒᵤₜ
  • rejected heat Qc
Sign conventions (cycle magnitudes used here)

We use positive per-cycle magnitudes:

  • Qₕ > 0 in from the hot reservoir.
  • Qc > 0 rejected to the cold reservoir.
  • Wₒᵤₜ > 0 produced.

First-law check: Wₒᵤₜ = Qₕ-Qc.

Second law (via Carnot bound) adds: e = Wₒᵤₜ/Qₕ ≤ 1-Tc/Tₕ.

Core derivation/explanation

From e ≤ 1-Tc/Tₕ, we get Wₒᵤₜ/Qₕ ≤ 1-Tc/Tₕ.

So even ideally, W_(out,max) = Qₕ(1-Tc/Tₕ).

Then minimum unavoidable rejection is Q_(c, min) = Qₕ-W_(out,max) = QₕTc/Tₕ.

This unavoidable Qc is the thermal bottleneck.

Units check: Tc/Tₕ is dimensionless (K/K), so multiplying by Qₕ still gives energy units (J).

Physical sanity/limit/sign check:

  • If Tc is close to Tₕ, bottleneck is severe and efficiency is low.
  • Lowering Tc or raising Tₕ can help, but never gives efficiency above 1.
  • As long as Tc > 0, some heat rejection is unavoidable.

Worked example(s)

Best-case output from a thermal plant

Worked example 1

Unavoidable rejected heat

Problem

A thermal plant receives 100 MJ of heat over a complete number of cycles between reservoirs at 900 K and 300 K. Find its maximum efficiency, maximum work output and minimum rejected heat over that period.
Show full solution

  1. Apply the temperature ceiling

    Method

    Use the Carnot bound rather than the first-law limit of complete conversion.

    Reason

    Finite reservoir temperatures require heat rejection even for a reversible engine.

    Working

    Answer

    eₘₐₓ = 1-300/900 = 0.667.

  2. Find the maximum work

    Method

    Multiply heat input by the maximum convertible fraction.

    Reason

    Efficiency is work output divided by heat input.

    Working

    Answer

    W_(out,max) = 0.667(100 MJ) ≈ 66.7 MJ.

  3. State the unavoidable remainder

    Working

    Answer

    Q_(c, min) = 100-66.7 = 33.3 MJ. The maximum efficiency is 66.7% and maximum work is 66.7 MJ. At least one-third of the heat input must be rejected; real losses increase that fraction.

Practice set

Check your understanding 1

Find the maximum engine efficiency between 800 K and 320 K reservoirs.

Show hint
Use the cold-to-hot absolute-temperature ratio.
Show answer
eₘₐₓ = 1-320/800 = 0.60 = 60%. Even the ideal cycle must reject the remaining 40% of its heat input.

Check your understanding 2

An engine receives 50 MJ per cycle and its maximum efficiency is 0.40. Find maximum work output and minimum rejected heat per cycle.

Show hint
First find the work fraction, then use the cycle energy balance.
Show answer
Wₘₐₓ = 0.40(50) = 20 MJ and Q_(c, min) = 50-20 = 30 MJ. These bounds refer to the ideal cycle at the stated efficiency ceiling.

Check your understanding 3

Can an engine between 700 K and 350 K reach 80% efficiency?

Show hint
Calculate the reservoir-dependent ceiling.
Show answer
No. eC = 1-350/700 = 0.50 = 50%, so 80% exceeds the second-law bound even though it is below 100%.

Check your understanding 4

Why does the second law create a thermal bottleneck for a cyclic power plant between finite positive temperatures?

Show hint
Consider the fraction that cannot become work.
Show answer
The plant must reject at least the fraction Tc/Tₕ of its heat input, leaving at most 1-Tc/Tₕ available as work.

Summary + next steps

  • First law says energy is conserved; second law sets conversion limits.
  • Carnot bound quantifies the thermal bottleneck directly from Tₕ and Tc.
  • Waste heat is not a design flaw alone; part of it is fundamentally unavoidable.

Previous: the Carnot engine Back to thermodynamics

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