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Carnot heat engine

Let heat flow between hot and cold reservoirs and understand the theoretical limit on efficiency.

ηₘₐₓ = 1 − T꜀/Tₕ W = Qₕ − Q꜀Senior high physics
Orange: heat in, green: heat out, blue: work
t = 0.00 s
Speed
Carnot efficiency limit57.1%
Work out per cycle285.7J
Heat to cold reservoir214.3J

Carnot efficiency vs. hot reservoir

75.037.50.0450563675788900ηₘₐₓ (%)Tₕ (K)

The cold-reservoir temperature uses the current setting; the Carnot efficiency is the ideal upper limit.

Think about it

With the hot reservoir unchanged, how does the engine's maximum efficiency change if the cold-reservoir temperature is raised?

Model assumptionsIdeal reversible cycle · Reservoir temperatures constant, in kelvin · No friction or heat loss · Heat input per cycle set by the parameter

Data log

Set the parameters and click “Record data”. Record several rows with different parameters to plot their relationship and fit a straight line; pause and record at different times to study how a reading changes with time.

How to use the lab

  1. Choose an experimentBrowse the catalog on the left by level or topic, or search for a concept.
  2. Change a parameterDrag a slider or type a value in range. The experiment restarts from its initial state right away.
  3. Slow down and observePause, change the speed, drag to look from another angle in 3D, and use the readings and graph to understand what changes.
  4. Test your ideasRead the explanation and the model assumptions, then answer the question under “Think about it”.