University Physics IV · topic UPIV-01.05
Heat capacities and the freezing out of degrees of freedom
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.
Heat capacities and frozen degrees of freedom: predict a finite change
A degree of freedom is frozen out when the quantum it must absorb, ħω, is far larger than the thermal energy k_BT available, so it stores almost no energy and contributes almost nothing to the heat capacity. Use the supplied topic model Y_out = 24.942*(f*d/t)^2*exp(-(f*d/t))/max(1-exp(-(f*d/t)),0.000000001)^2, where Y_out is EINSTEIN C_V (J mol⁻¹ K⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. d is Debye temperature θ_D, measured in K. Keep f (Einstein ratio θ_E / θ_D) = 0.75; t (Temperature T) = 20 K fixed. For this exercise use only 142.5 K ≤ d ≤ 457.5 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 d = 211.8 K and 255.9 K. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EINSTEIN C_V against d
Predictions from the supplied Heat capacities and frozen degrees of freedom 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): EINSTEIN C_V / J mol⁻¹ K⁻¹
Horizontal axis (x): d: Debye temperature θ_D / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Debye temperature θ_D / K | EINSTEIN C_V / J mol⁻¹ K⁻¹ |
|---|---|
| 142.5 | 3.4357566505694526 |
| 221.25 | 0.42824263352654585 |
| 300 | 0.041061489230309334 |
| 378.75 | 0.0034147653785792003 |
| 457.5 | 0.00025996287586963314 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 211.8 K, substitute into Y_out = 24.942*(f*d/t)^2*exp(-(f*d/t))/max(1-exp(-(f*d/t)),0.000000001)^2: Y₀ = 0.559462 J mol⁻¹ K⁻¹.
- At d = 255.9 K, the same fixed-parameter model gives Y₁ = 0.156169 J mol⁻¹ K⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.403293 J mol⁻¹ K⁻¹; the input change is Δd = 44.1 K.
- Average sensitivity is ΔY_out/Δd = -0.00914496 J mol⁻¹ K⁻¹ per K. Its negative sign means that this response decreases 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 = 0.559462 J mol⁻¹ K⁻¹, Y₁ = 0.156169 J mol⁻¹ K⁻¹; ΔY_out = -0.403293 J mol⁻¹ K⁻¹; average sensitivity = -0.00914496 J mol⁻¹ 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.
Heat capacities and frozen degrees of freedom: infer an input with uncertainty
A degree of freedom is frozen out when the quantum it must absorb, ħω, is far larger than the thermal energy k_BT available, so it stores almost no energy and contributes almost nothing to the heat capacity. Use the supplied topic model Y_out = 24.942*(f*d/t)^2*exp(-(f*d/t))/max(1-exp(-(f*d/t)),0.000000001)^2, where Y_out is EINSTEIN C_V (J mol⁻¹ K⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ; max(a,b) selects the larger value. d is Debye temperature θ_D, measured in K. Keep f (Einstein ratio θ_E / θ_D) = 0.75; t (Temperature T) = 20 K fixed. For this exercise use only 142.5 K ≤ d ≤ 457.5 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 = 0.013933308 J mol⁻¹ K⁻¹, with a bounded reading uncertainty ±0.002051 J mol⁻¹ 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 d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0315 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.
EINSTEIN C_V against d
Predictions from the supplied Heat capacities and frozen degrees of freedom 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): EINSTEIN C_V / J mol⁻¹ K⁻¹
Horizontal axis (x): d: Debye temperature θ_D / K
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Debye temperature θ_D / K | EINSTEIN C_V / J mol⁻¹ K⁻¹ |
|---|---|
| 142.5 | 3.4357566505694526 |
| 221.25 | 0.42824263352654585 |
| 300 | 0.041061489230309334 |
| 378.75 | 0.0034147653785792003 |
| 457.5 | 0.00025996287586963314 |
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(d) = 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(142.5) = 3.43576 J mol⁻¹ K⁻¹ and Y_out(457.5) = 0.000259963 J mol⁻¹ K⁻¹. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 142.5 + (0.0139333 − 3.43576)(457.5 − 142.5)/(0.000259963 − 3.43576) = 456.246 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 d = 334.65 K; substituting back gives Y_out = 0.013933308 J mol⁻¹ K⁻¹.
- The interpolation error in the input is |d_line − d| = 121.596 K. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.011882308 J mol⁻¹ K⁻¹ and 0.015984308 J mol⁻¹ K⁻¹. Sort the roots even if the response decreases: 330.288 K ≤ d ≤ 339.694 K.
- The lower and upper input deviations are 4.36166 K and 5.04401 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 d ≈ 334.65 K; compatible interval [330.288, 339.694] K. Endpoint interpolation gives 456.246 K.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.