University Physics II · topic UPII-E2.01
Temperature and thermal equilibrium
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.
Temperature and thermal equilibrium: predict a finite change
Temperature is the single number shared by every system in thermal equilibrium with a given body, so that two systems at the same temperature exchange no net energy when they are placed in thermal contact. Use the supplied topic model Y_out = p0*(tc+273.15)/273.15, where Y_out is PRESSURE p (×10⁵ Pa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p0 is Ice-point pressure p₀, measured in ×10⁵ Pa. Keep tc (Bath temperature t) = 100 °C fixed. For this exercise use only 0.70625 ×10⁵ Pa ≤ p0 ≤ 1.49375 ×10⁵ Pa 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 p0 = 0.8795 ×10⁵ Pa and 0.98975 ×10⁵ Pa. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
PRESSURE p against p0
Predictions from the supplied Temperature and thermal equilibrium 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): PRESSURE p / ×10⁵ Pa
Horizontal axis (x): p0: Ice-point pressure p₀ / ×10⁵ Pa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p0: Ice-point pressure p₀ / ×10⁵ Pa | PRESSURE p / ×10⁵ Pa |
|---|---|
| 0.7062499999999999 | 0.9648075691012263 |
| 0.903125 | 1.233758351638294 |
| 1.0999999999999999 | 1.5027091341753613 |
| 1.296875 | 1.771659916712429 |
| 1.49375 | 2.040610699249496 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p0 = 0.8795 ×10⁵ Pa, substitute into Y_out = p0*(tc+273.15)/273.15: Y₀ = 1.20148 ×10⁵ Pa.
- At p0 = 0.98975 ×10⁵ Pa, the same fixed-parameter model gives Y₁ = 1.3521 ×10⁵ Pa.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.150612 ×10⁵ Pa; the input change is Δp0 = 0.11025 ×10⁵ Pa.
- Average sensitivity is ΔY_out/Δp0 = 1.3661 ×10⁵ Pa per ×10⁵ Pa. 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 = 1.20148 ×10⁵ Pa, Y₁ = 1.3521 ×10⁵ Pa; ΔY_out = 0.150612 ×10⁵ Pa; average sensitivity = 1.3661 ×10⁵ Pa per ×10⁵ Pa.
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.
Temperature and thermal equilibrium: infer an input with uncertainty
Temperature is the single number shared by every system in thermal equilibrium with a given body, so that two systems at the same temperature exchange no net energy when they are placed in thermal contact. Use the supplied topic model Y_out = p0*(tc+273.15)/273.15, where Y_out is PRESSURE p (×10⁵ Pa). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p0 is Ice-point pressure p₀, measured in ×10⁵ Pa. Keep tc (Bath temperature t) = 100 °C fixed. For this exercise use only 0.70625 ×10⁵ Pa ≤ p0 ≤ 1.49375 ×10⁵ Pa 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 = 1.6210475 ×10⁵ Pa, with a bounded reading uncertainty ±0.062934 ×10⁵ Pa. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875 ×10⁵ Pa, 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.
PRESSURE p against p0
Predictions from the supplied Temperature and thermal equilibrium 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): PRESSURE p / ×10⁵ Pa
Horizontal axis (x): p0: Ice-point pressure p₀ / ×10⁵ Pa
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p0: Ice-point pressure p₀ / ×10⁵ Pa | PRESSURE p / ×10⁵ Pa |
|---|---|
| 0.7062499999999999 | 0.9648075691012263 |
| 0.903125 | 1.233758351638294 |
| 1.0999999999999999 | 1.5027091341753613 |
| 1.296875 | 1.771659916712429 |
| 1.49375 | 2.040610699249496 |
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(p0) = 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(0.70625) = 0.964808 ×10⁵ Pa and Y_out(1.49375) = 2.04061 ×10⁵ Pa. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives p0_line = 0.70625 + (1.62105 − 0.964808)(1.49375 − 0.70625)/(2.04061 − 0.964808) = 1.18663 ×10⁵ Pa. 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 p0 = 1.18663 ×10⁵ Pa; substituting back gives Y_out = 1.6210475 ×10⁵ Pa.
- The interpolation error in the input is |p0_line − p0| = 4.44089e-16 ×10⁵ Pa. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.5581135 ×10⁵ Pa and 1.6839815 ×10⁵ Pa. Sort the roots even if the response decreases: 1.14056 ×10⁵ Pa ≤ p0 ≤ 1.23269 ×10⁵ Pa.
- The lower and upper input deviations are 0.0460684 ×10⁵ Pa and 0.0460684 ×10⁵ Pa. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer p0 ≈ 1.18663 ×10⁵ Pa; compatible interval [1.14056, 1.23269] ×10⁵ Pa. Endpoint interpolation gives 1.18663 ×10⁵ Pa.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.