Thermal physics
Heat engines, efficiency, and the Carnot cycle
Explore the ideal efficiency of a heat engine between a hot reservoir Th and a cold reservoir Tc. Read ηCarnot, maximum work Wmax, and rejected heat Qc; verify and .
⚠ For real experiments with hot steam, heated gas, or pressure vessels, wear protection, never open a pressurized container, and keep a safe distance from high-temperature sources.
Equipment
- Canvas showing the closed p–V cycle and the engine's heat-flow diagram
- «Nguồn nóng Th» (hot reservoir) slider, 350–900 K
- «Nguồn lạnh Tc» (cold reservoir) slider, 180–340 K
- «Nhiệt nhận Qh» (absorbed heat) slider, 100–1000 J
- Readout line «ηCarnot = …; Wmax = ηQh = …; Qc = Qh − W = …»
Procedure
Measure efficiency vs hot temperature
Hold «Nguồn lạnh Tc» at 300 K, raise «Nguồn nóng Th» from 350 to 900 K, and record «ηCarnot» at each step. Compute for each pair and compare with the readout; watch the efficiency approach 100 % without ever reaching it.
Vary the cold reservoir and absorbed heat
With Th = 600 K, slide «Nguồn lạnh Tc» from 180 to 340 K: efficiency falls as the reservoirs approach each other. Then fix Th and Tc, change «Nhiệt nhận Qh», and confirm η stays the same — only «Wmax = ηQh» and «Qc = Qh − W» scale with Qh. This shows depends only on the temperatures, not on the size of the cycle.
Read the cycle on the p–V diagram
Look at the shaded closed loop on the canvas: the «chiều chu trình» (cycle direction) arrow indicates the forward (clockwise) run of an engine, and the enclosed area is proportional to the net work W. Drag the figure to change the viewpoint. Relate the Wmax readout to the area interpretation: is positive for a clockwise cycle.
The second-law limit
Drag Tc close to Th: the efficiency drops toward 0 — no work can be extracted when the reservoirs share a temperature. Conversely set Th = 900 K and Tc = 180 K (the slider limits): η ≈ 80 % is still below 100 %. Predict that η → 1 would require Tc → 0 K, which is unattainable — that is precisely the second-law bound.