University Physics II · unit UPII-E2
Institutional Extension · Temperature, Heat, and Kinetic Theory
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
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.
Heat & Specific Heat: infer an input with uncertainty
The specific heat capacity c of a material is the energy that must be transferred to raise one kilogram of it by one kelvin with no change of phase; heat Q is that transferred energy itself, measured in joules. Use the supplied topic model Y_out = (m*c*t0 + 837.2*20)/(m*c + 837.2), where Y_out is FINAL TEMPERATURE T_f (°C). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. m is Metal mass m, measured in kg. Keep c (Metal specific heat c) = 400 J kg⁻¹ K⁻¹; t0 (Metal start temperature) = 95 °C fixed. For this exercise use only 0.150938 kg ≤ m ≤ 0.899062 kg 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 = 36.867436 °C, with a bounded reading uncertainty ±0.85031 °C. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125 kg, 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.
FINAL TEMPERATURE T_f against m
Predictions from the supplied Heat & Specific Heat 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): FINAL TEMPERATURE T_f / °C
Horizontal axis (x): m: Metal mass m / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Metal mass m / kg | FINAL TEMPERATURE T_f / °C |
|---|---|
| 0.1509375 | 25.04484304932735 |
| 0.33796875 | 30.426977413839644 |
| 0.5249999999999999 | 35.04010695187166 |
| 0.7120312499999999 | 39.038056617016295 |
| 0.8990624999999999 | 42.53618950138908 |
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(m) = 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(0.150938) = 25.0448 °C and Y_out(0.899062) = 42.5362 °C. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 0.150938 + (36.8674 − 25.0448)(0.899062 − 0.150938)/(42.5362 − 25.0448) = 0.656603 kg. 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 m = 0.607294 kg; substituting back gives Y_out = 36.867436 °C.
- The interpolation error in the input is |m_line − m| = 0.0493096 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 36.017126 °C and 37.717746 °C. Sort the roots even if the response decreases: 0.568366 kg ≤ m ≤ 0.647377 kg.
- The lower and upper input deviations are 0.038928 kg and 0.0400837 kg. 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 m ≈ 0.607294 kg; compatible interval [0.568366, 0.647377] kg. Endpoint interpolation gives 0.656603 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Latent heat and phase changes: infer an input with uncertainty
The specific latent heat of a substance is the energy that must be supplied per kilogram to carry it through a phase change at constant temperature, and that is given back per kilogram when the change runs the other way. Use the supplied topic model Y_out = lf/cliq, where Y_out is PLATEAU IN DEGREES L_f/c (K). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. cliq is Specific heat of the liquid c, measured in kJ kg⁻¹ K⁻¹. Keep lf (Latent heat of fusion L_f) = 334 kJ kg⁻¹; q (Energy supplied Q) = 300 kJ fixed. For this exercise use only 2.31875 kJ kg⁻¹ K⁻¹ ≤ cliq ≤ 4.68125 kJ kg⁻¹ 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 = 88.83274 K, with a bounded reading uncertainty ±2.6226 K. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown cliq by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 kJ kg⁻¹ 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.
PLATEAU IN DEGREES L_f/c against cliq
Predictions from the supplied Latent heat and phase 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): PLATEAU IN DEGREES L_f/c / K
Horizontal axis (x): cliq: Specific heat of the liquid c / kJ kg⁻¹ K⁻¹
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| cliq: Specific heat of the liquid c / kJ kg⁻¹ K⁻¹ | PLATEAU IN DEGREES L_f/c / K |
|---|---|
| 2.31875 | 144.04312668463612 |
| 2.9093750000000003 | 114.80128893662727 |
| 3.5 | 95.42857142857143 |
| 4.090625 | 81.65011459129106 |
| 4.68125 | 71.34846461949265 |
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(cliq) = 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(2.31875) = 144.043 K and Y_out(4.68125) = 71.3485 K. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives cliq_line = 2.31875 + (88.8327 − 144.043)(4.68125 − 2.31875)/(71.3485 − 144.043) = 4.11303 kJ kg⁻¹ 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 cliq = 3.75988 kJ kg⁻¹ K⁻¹; substituting back gives Y_out = 88.83274 K.
- The interpolation error in the input is |cliq_line − cliq| = 0.353154 kJ kg⁻¹ K⁻¹. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 86.21014 K and 91.45534 K. Sort the roots even if the response decreases: 3.65206 kJ kg⁻¹ K⁻¹ ≤ cliq ≤ 3.87425 kJ kg⁻¹ K⁻¹.
- The lower and upper input deviations are 0.107819 kJ kg⁻¹ K⁻¹ and 0.114379 kJ kg⁻¹ K⁻¹. 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 cliq ≈ 3.75988 kJ kg⁻¹ K⁻¹; compatible interval [3.65206, 3.87425] kJ kg⁻¹ K⁻¹. Endpoint interpolation gives 4.11303 kJ kg⁻¹ K⁻¹.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.