Differential Equation Model of Cooling with a Changed Ambient
A cup of coffee cools in a room. Newton's law of cooling gives dT/dt = −k(T − Tₐ), where T is the coffee temperature in °C, t is in minutes, and Tₐ is the ambient temperature. Initially T = 90 °C and Tₐ = 20 °C, and after 10 minutes the coffee is 65 °C. At t = 10 the room's heating switches on so that Tₐ becomes 30 °C for all later times. A student solves the first stage correctly, obtaining k ≈ 0.0513, then applies the same formula T = 20 + 70e^{−kt} for t > 10. The marking guide requires correct treatment of the piecewise ambient temperature and continuity at t = 10.
1. Find k for the first 10 minutes and the temperature at t = 10, then set up and solve the model for t > 10.
Read the worked answer
For 0 ≤ t ≤ 10: T = 20 + 70e^{−kt}. At t = 10, 65 = 20 + 70e^{−10k}, so e^{−10k} = 45/70 = 9/14, giving k = −(1/10)ln(9/14) ≈ 0.0442 min⁻¹. (Note: 0.0513 arises from an incorrect initial difference; the correct initial excess is 70, not 90 − 20 computed as 70 but then misused.)
At t = 10 the temperature is 65 °C, which is continuous.
For t > 10, Tₐ = 30, so dT/dt = −k(T − 30). The solution is T = 30 + Ae^{−kt}. Continuity at t = 10 requires 65 = 30 + Ae^{−10k}, so A = 35e^{10k}. Since e^{−10k} = 9/14, e^{10k} = 14/9, so A = 35(14/9) ≈ 54.44. Thus T = 30 + 54.44e^{−0.0442(t−10)} for t ≥ 10.
2. Determine the long-term temperature and the time after t = 10 at which the coffee reaches 40 °C. Explain why reusing the original formula gives a wrong answer.
Read the worked answer
As t → ∞, e^{−k(t−10)} → 0, so T → 30 °C, the new ambient temperature. This is physically sensible: the coffee can never cool below the surrounding air temperature.
Set 40 = 30 + 54.44e^{−0.0442(t−10)}: e^{−0.0442(t−10)} = 10/54.44 ≈ 0.1837, so −0.0442(t − 10) = ln(0.1837) ≈ −1.694, giving t − 10 ≈ 38.3 min, so t ≈ 48.3 min.
Reusing T = 20 + 70e^{−kt} is wrong because it embeds the old ambient temperature of 20 °C as the asymptote and the original initial excess of 70. After t = 10 the driving term has changed to (T − 30), so the equilibrium shifts and the constant must be recalibrated using the temperature at t = 10. The exponential decay rate k is unchanged because it depends on the cooling properties, not the ambient value.