University Physics II · topic UPII-E3.09
Calculating entropy changes
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.
Calculating entropy changes: predict a finite change
The entropy change between two equilibrium states is the integral of dQ/T along any reversible path joining them — a number fixed by the two states alone, not by the process that actually happened. Use the supplied topic model Y_out = m*4186*ln(tf/273.15), where Y_out is HEATING ΔS (J K⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. tf is Final temperature T_f, measured in K. Keep m (Mass of ice m) = 0.7 kg fixed. For this exercise use only 294.031 K ≤ tf ≤ 360.969 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 tf = 308.757 K and 318.129 K. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
HEATING ΔS against tf
Predictions from the supplied Calculating entropy changes 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): HEATING ΔS / J K⁻¹
Horizontal axis (x): tf: Final temperature T_f / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tf: Final temperature T_f / K | HEATING ΔS / J K⁻¹ |
|---|---|
| 294.03125 | 215.85306393223857 |
| 310.765625 | 378.0482970583869 |
| 327.5 | 531.7344868764364 |
| 344.234375 | 677.7601042393605 |
| 360.96875 | 816.852717375305 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of tf separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At tf = 308.757 K, substitute into Y_out = m*4186*ln(tf/273.15): Y₀ = 359.052 J K⁻¹.
- At tf = 318.129 K, the same fixed-parameter model gives Y₁ = 446.665 J K⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 87.613 J K⁻¹; the input change is Δtf = 9.37125 K.
- Average sensitivity is ΔY_out/Δtf = 9.34912 J K⁻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 = 359.052 J K⁻¹, Y₁ = 446.665 J K⁻¹; ΔY_out = 87.613 J K⁻¹; average sensitivity = 9.34912 J K⁻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.
Calculating entropy changes: infer an input with uncertainty
The entropy change between two equilibrium states is the integral of dQ/T along any reversible path joining them — a number fixed by the two states alone, not by the process that actually happened. Use the supplied topic model Y_out = m*4186*ln(tf/273.15), where Y_out is HEATING ΔS (J K⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; ln(x) means the natural logarithm. tf is Final temperature T_f, measured in K. Keep m (Mass of ice m) = 0.7 kg fixed. For this exercise use only 294.031 K ≤ tf ≤ 360.969 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 = 596.884 J K⁻¹, with a bounded reading uncertainty ±32.995 J K⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown tf by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00669375 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.
HEATING ΔS against tf
Predictions from the supplied Calculating entropy changes 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): HEATING ΔS / J K⁻¹
Horizontal axis (x): tf: Final temperature T_f / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| tf: Final temperature T_f / K | HEATING ΔS / J K⁻¹ |
|---|---|
| 294.03125 | 215.85306393223857 |
| 310.765625 | 378.0482970583869 |
| 327.5 | 531.7344868764364 |
| 344.234375 | 677.7601042393605 |
| 360.96875 | 816.852717375305 |
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(tf) = 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(294.031) = 215.853 J K⁻¹ and Y_out(360.969) = 816.853 J K⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives tf_line = 294.031 + (596.884 − 215.853)(360.969 − 294.031)/(816.853 − 215.853) = 336.469 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 tf = 334.863 K; substituting back gives Y_out = 596.884 J K⁻¹.
- The interpolation error in the input is |tf_line − tf| = 1.60618 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 563.889 J K⁻¹ and 629.879 J K⁻¹. Sort the roots even if the response decreases: 331.114 K ≤ tf ≤ 338.655 K.
- The lower and upper input deviations are 3.74952 K and 3.79198 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 tf ≈ 334.863 K; compatible interval [331.114, 338.655] K. Endpoint interpolation gives 336.469 K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.