The Cooling Engine Model
An engineer models the temperature T (°C) of an engine t minutes after shutdown using T = 20 + 80e^(−kt). At t = 0 the temperature is 100 °C, consistent with the model. After 5 minutes the temperature is 60 °C. The engineer wants to know the cooling constant k, the time for the engine to cool to 30 °C, and the rate of cooling at t = 5 minutes. A colleague suggests taking logarithms of both sides of 40 = 80e^(−5k) and warns that the rate of cooling is not simply the difference in temperatures divided by the time.
1. Find k and the time at which the engine reaches 30 °C, showing each logarithmic step.
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At t = 5: 60 = 20 + 80e^(−5k) → 40 = 80e^(−5k) → e^(−5k) = 0.5. Taking natural logs: −5k = ln 0.5 = −ln 2, so k = (ln 2)/5 ≈ 0.1386 per minute.
For T = 30: 30 = 20 + 80e^(−kt) → 10 = 80e^(−kt) → e^(−kt) = 0.125. So −kt = ln 0.125 = −ln 8, giving t = (ln 8)/k = (3 ln 2)/((ln 2)/5) = 15 minutes. The engine reaches 30 °C after 15 minutes.
2. Compute the instantaneous rate of cooling at t = 5 minutes and compare it with the average rate over the first 5 minutes.
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Differentiate: dT/dt = −80k·e^(−kt). At t = 5, e^(−5k) = 0.5, so dT/dt = −80(0.1386)(0.5) ≈ −5.54 °C per minute. The negative sign shows temperature is falling.
The average rate over 0 to 5 minutes is (60 − 100)/5 = −8 °C per minute. The instantaneous rate at t = 5 (−5.54 °C/min) is smaller in magnitude than the average, which is expected for exponential decay: cooling is fastest at the start and slows over time. The colleague is right that dividing the temperature change by the time gives only an average, not the rate at a specific instant.