University Physics II · topic UPII-E2.02
Thermal expansion
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.
Thermal Expansion & Thermal Stress: predict a finite change
Thermal expansion is the fractional change in a body's dimensions per kelvin of temperature change. The linear coefficient α = (1/L)(dL/dT) has units K⁻¹, and under uniform heating every length grows by the same factor 1 + α ΔT. Use the supplied topic model Y_out = al*L0*dT/1000, where Y_out is GROWTH ΔL = α L₀ ΔT (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. al is Coefficient α (steel 12, aluminium 23), measured in × 10⁻⁶ K⁻¹. Keep dT (Temperature rise ΔT) = 40 K; L0 (Bar length L₀) = 2 m fixed. For this exercise use only 4.08125 × 10⁻⁶ K⁻¹ ≤ al ≤ 26.9187 × 10⁻⁶ K⁻¹ 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 al = 9.1055 × 10⁻⁶ K⁻¹ and 12.3027 × 10⁻⁶ K⁻¹. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δal across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
GROWTH ΔL = α L₀ ΔT against al
Predictions from the supplied Thermal Expansion & Thermal Stress 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): GROWTH ΔL = α L₀ ΔT / mm
Horizontal axis (x): al: Coefficient α (steel 12, aluminium 23) / × 10⁻⁶ K⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| al: Coefficient α (steel 12, aluminium 23) / × 10⁻⁶ K⁻¹ | GROWTH ΔL = α L₀ ΔT / mm |
|---|---|
| 4.08125 | 0.3265 |
| 9.790624999999999 | 0.7832499999999999 |
| 15.5 | 1.24 |
| 21.209374999999998 | 1.6967499999999998 |
| 26.91875 | 2.1535 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of al separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At al = 9.1055 × 10⁻⁶ K⁻¹, substitute into Y_out = al*L0*dT/1000: Y₀ = 0.72844 mm.
- At al = 12.3027 × 10⁻⁶ K⁻¹, the same fixed-parameter model gives Y₁ = 0.98422 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.25578 mm; the input change is Δal = 3.19725 × 10⁻⁶ K⁻¹.
- Average sensitivity is ΔY_out/Δal = 0.08 mm per × 10⁻⁶ K⁻¹. 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 = 0.72844 mm, Y₁ = 0.98422 mm; ΔY_out = 0.25578 mm; average sensitivity = 0.08 mm per × 10⁻⁶ K⁻¹.
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.
Thermal Expansion & Thermal Stress: infer an input with uncertainty
Thermal expansion is the fractional change in a body's dimensions per kelvin of temperature change. The linear coefficient α = (1/L)(dL/dT) has units K⁻¹, and under uniform heating every length grows by the same factor 1 + α ΔT. Use the supplied topic model Y_out = al*L0*dT/1000, where Y_out is GROWTH ΔL = α L₀ ΔT (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. al is Coefficient α (steel 12, aluminium 23), measured in × 10⁻⁶ K⁻¹. Keep dT (Temperature rise ΔT) = 40 K; L0 (Bar length L₀) = 2 m fixed. For this exercise use only 4.08125 × 10⁻⁶ K⁻¹ ≤ al ≤ 26.9187 × 10⁻⁶ K⁻¹ 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 = 1.44097 mm, with a bounded reading uncertainty ±0.10688 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown al by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00228375 × 10⁻⁶ K⁻¹, 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.
GROWTH ΔL = α L₀ ΔT against al
Predictions from the supplied Thermal Expansion & Thermal Stress 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): GROWTH ΔL = α L₀ ΔT / mm
Horizontal axis (x): al: Coefficient α (steel 12, aluminium 23) / × 10⁻⁶ K⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| al: Coefficient α (steel 12, aluminium 23) / × 10⁻⁶ K⁻¹ | GROWTH ΔL = α L₀ ΔT / mm |
|---|---|
| 4.08125 | 0.3265 |
| 9.790624999999999 | 0.7832499999999999 |
| 15.5 | 1.24 |
| 21.209374999999998 | 1.6967499999999998 |
| 26.91875 | 2.1535 |
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(al) = 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(4.08125) = 0.3265 mm and Y_out(26.9187) = 2.1535 mm. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives al_line = 4.08125 + (1.44097 − 0.3265)(26.9187 − 4.08125)/(2.1535 − 0.3265) = 18.0121 × 10⁻⁶ K⁻¹. 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 al = 18.0121 × 10⁻⁶ K⁻¹; substituting back gives Y_out = 1.44097 mm.
- The interpolation error in the input is |al_line − al| = 0 × 10⁻⁶ K⁻¹. This is within the requested numerical tolerance.
- Invert both limits Y_out = 1.33409 mm and 1.54785 mm. Sort the roots even if the response decreases: 16.6761 × 10⁻⁶ K⁻¹ ≤ al ≤ 19.3481 × 10⁻⁶ K⁻¹.
- The lower and upper input deviations are 1.336 × 10⁻⁶ K⁻¹ and 1.336 × 10⁻⁶ K⁻¹. 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 al ≈ 18.0121 × 10⁻⁶ K⁻¹; compatible interval [16.6761, 19.3481] × 10⁻⁶ K⁻¹. Endpoint interpolation gives 18.0121 × 10⁻⁶ K⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.