Modelling a Cooling Cup
A science class measures the temperature of a cup of coffee left in a room kept at 21°C. Temperature T (°C) is recorded every 5 minutes. The data show a rapid initial drop that slows and levels off. A student proposes a linear model T = a + bt. A second student proposes an exponential-decay-to-ambient model T = 21 + A·e^(−kt). The measurements are: t = 0, T = 88.0; t = 5, T = 71.5; t = 10, T = 59.0; t = 15, T = 50.0; t = 20, T = 43.5; t = 25, T = 38.5; t = 30, T = 35.0. The class must choose a model and use it to predict when the coffee reaches 30°C.
1. Using the transformed variable y = T − 21, show that the exponential model is appropriate by checking whether y decays by a roughly constant ratio per 5-minute interval. Estimate A and k.
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Compute y = T − 21: 67.0, 50.5, 38.0, 29.0, 22.5, 17.5, 14.0.\n\nSuccessive ratios: 50.5/67.0 ≈ 0.754; 38.0/50.5 ≈ 0.752; 29.0/38.0 ≈ 0.763; 22.5/29.0 ≈ 0.776; 17.5/22.5 ≈ 0.778; 14.0/17.5 ≈ 0.800.\n\nThe ratios are roughly constant (about 0.75–0.80), which supports exponential decay of y. The slight drift upward may reflect measurement error or a small ambient-temperature error.\n\nA is the value of y at t = 0: A = 67.0.\n\nUsing the 5-minute ratio r ≈ 0.754: e^(−5k) = 0.754, so −5k = ln 0.754 ≈ −0.2823, giving k ≈ 0.0565 per minute.\n\nCheck at t = 20: y = 67.0 × e^(−0.0565 × 20) = 67.0 × e^(−1.13) ≈ 67.0 × 0.323 ≈ 21.6, so T ≈ 42.6°C, close to the measured 43.5°C. The model fits well.
2. Use the exponential model to predict when the coffee reaches 30°C, and explain why the linear model fails for this prediction.
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Set T = 30, so y = 9.0.\n9.0 = 67.0 × e^(−0.0565t).\ne^(−0.0565t) = 9.0/67.0 ≈ 0.1343.\n−0.0565t = ln 0.1343 ≈ −2.008.\nt ≈ 2.008/0.0565 ≈ 35.5 minutes.\n\nSo the coffee reaches 30°C at about 35–36 minutes after the start.\n\nWhy the linear model fails: a linear model T = a + bt fitted to the early data would have a steep negative slope (roughly −1.8°C per minute from the first intervals). Extrapolating, it predicts T = 30 at about t ≈ 32 minutes - not far off numerically here - but it continues downward indefinitely, predicting 0°C at around 49 minutes and negative temperatures later. That is physically impossible: the coffee cannot cool below the 21°C room. The exponential model has a horizontal asymptote at 21°C, matching the physics. The linear model also fits the data poorly overall (the curve is clearly concave up), so its predictions are unreliable outside the measured range.
3. The class later realises the room temperature was actually 23°C, not 21°C. Explain how this changes the model and whether the prediction of the time to reach 30°C changes much.
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With ambient 23°C, define y = T − 23: 65.0, 48.5, 36.0, 27.0, 20.5, 15.5, 12.0.\nA = 65.0.\nRatios: 48.5/65.0 ≈ 0.746; 36.0/48.5 ≈ 0.742; 27.0/36.0 ≈ 0.750; 20.5/27.0 ≈ 0.759; 15.5/20.5 ≈ 0.756; 12.0/15.5 ≈ 0.774.\nThe ratios are now more consistent (about 0.75), so the corrected ambient temperature actually improves the fit - evidence that 23°C is the better value.\n\ne^(−5k) = 0.75 gives k ≈ 0.0575 per minute.\n\nTime to reach 30°C: y = 30 − 23 = 7.0.\n7.0 = 65.0 e^(−0.0575t) → e^(−0.0575t) = 0.1077 → t = ln(0.1077)/(−0.0575) ≈ 2.228/0.0575 ≈ 38.7 minutes.\n\nThe prediction shifts from about 35.5 to about 38.7 minutes - roughly 3 minutes later. The judgement point: the asymptote strongly affects predictions near the ambient temperature, so getting the ambient value right matters more for long-range predictions than for the early fit. The corrected model is more trustworthy overall.