University Physics II · topic UPII-E2.03
Heat and specific heat
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 & Specific Heat: predict a finite change
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. Calculate Y_out at m = 0.315525 kg and 0.420262 kg. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of m separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m = 0.315525 kg, substitute into Y_out = (m*c*t0 + 837.2*20)/(m*c + 837.2): Y₀ = 29.8253 °C.
- At m = 0.420262 kg, the same fixed-parameter model gives Y₁ = 32.5413 °C.
- Signed change: ΔY_out = Y₁ − Y₀ = 2.71609 °C; the input change is Δm = 0.104737 kg.
- Average sensitivity is ΔY_out/Δm = 25.9323 °C per kg. 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 = 29.8253 °C, Y₁ = 32.5413 °C; ΔY_out = 2.71609 °C; average sensitivity = 25.9323 °C per kg.
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 & 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.