University Physics II · topic UPII-E3.05
The Carnot cycle and maximum efficiency
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
The Carnot cycle and maximum efficiency: predict a finite change
The Carnot cycle is the reversible cycle built from two isothermal stages, one at each reservoir temperature, joined by two adiabatic stages, and its efficiency 1 − T_C/T_H is the highest any engine working between those reservoirs can reach. Use the supplied topic model Y_out = 1-tc/th, where Y_out is CARNOT EFFICIENCY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. th is Hot reservoir T_H, measured in K. Keep tc (Cold reservoir T_C) = 300 K; r (Expansion ratio V_b/V_a) = 2.4 fixed. For this exercise use only 503.375 K ≤ th ≤ 676.625 K on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at th = 541.49 K and 565.745 K. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δth across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
CARNOT EFFICIENCY against th
Predictions from the supplied The Carnot cycle and maximum efficiency model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CARNOT EFFICIENCY
Horizontal axis (x): th: Hot reservoir T_H / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Hot reservoir T_H / K | CARNOT EFFICIENCY |
|---|---|
| 503.375 | 0.4040228457909113 |
| 546.6875 | 0.45124042528867037 |
| 590 | 0.4915254237288136 |
| 633.3125 | 0.5263002072436593 |
| 676.625 | 0.5566229447626085 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of th separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At th = 541.49 K, substitute into Y_out = 1-tc/th: Y₀ = 0.445973.
- At th = 565.745 K, the same fixed-parameter model gives Y₁ = 0.469726.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0237526; the input change is Δth = 24.255 K.
- Average sensitivity is ΔY_out/Δth = 0.000979287 1 per K. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.445973, Y₁ = 0.469726; ΔY_out = 0.0237526; average sensitivity = 0.000979287 1 per K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
The Carnot cycle and maximum efficiency: infer an input with uncertainty
The Carnot cycle is the reversible cycle built from two isothermal stages, one at each reservoir temperature, joined by two adiabatic stages, and its efficiency 1 − T_C/T_H is the highest any engine working between those reservoirs can reach. Use the supplied topic model Y_out = 1-tc/th, where Y_out is CARNOT EFFICIENCY (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. th is Hot reservoir T_H, measured in K. Keep tc (Cold reservoir T_C) = 300 K; r (Expansion ratio V_b/V_a) = 2.4 fixed. For this exercise use only 503.375 K ≤ th ≤ 676.625 K on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 0.50743567, with a bounded reading uncertainty ±0.0073781. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.017325 K, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
CARNOT EFFICIENCY against th
Predictions from the supplied The Carnot cycle and maximum efficiency model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): CARNOT EFFICIENCY
Horizontal axis (x): th: Hot reservoir T_H / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Hot reservoir T_H / K | CARNOT EFFICIENCY |
|---|---|
| 503.375 | 0.4040228457909113 |
| 546.6875 | 0.45124042528867037 |
| 590 | 0.4915254237288136 |
| 633.3125 | 0.5263002072436593 |
| 676.625 | 0.5566229447626085 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(th) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(503.375) = 0.404023 and Y_out(676.625) = 0.556623. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 503.375 + (0.507436 − 0.404023)(676.625 − 503.375)/(0.556623 − 0.404023) = 620.782 K. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives th = 609.058 K; substituting back gives Y_out = 0.50743567.
- The interpolation error in the input is |th_line − th| = 11.7242 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.50005757 and 0.51481377. Sort the roots even if the response decreases: 600.069 K ≤ th ≤ 618.319 K.
- The lower and upper input deviations are 8.98841 K and 9.26178 K. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer th ≈ 609.058 K; compatible interval [600.069, 618.319] K. Endpoint interpolation gives 620.782 K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.