A cup of hot coffee is placed on a table to cool. Its temperature, \(T\) (°C), after \(t\) minutes is modelled by the function:
\(T(t) = 65e^{-0.08t} + 20\), for \(t \geq 0\).
(a) Write down the initial temperature of the coffee. [1]
(b) Find the temperature of the coffee after 10 minutes. [2]
(c) Find the time it takes for the coffee to cool down to 40 °C. [3]
(d) Write down the temperature of the room, according to the model. Justify your answer. [2]
(e) Find the rate at which the coffee is cooling after 5 minutes. [3]
(f) Find \(T'(t)\). [2]
(g) Explain what \(T'(t)\) represents in the context of this model. [1]
Question
Write down the initial temperature of the coffee. [1]
Hint
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Substitute t=0 into the function: T(0)=65e^(0)+20.
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T(0) = 85 °C
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Find the temperature of the coffee after 10 minutes. [2]
Hint
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Substitute t=10 into the model and calculate the exponential term.
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T(10) ≈ 49.2 °C
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Find the time it takes for the coffee to cool down to 40 °C. [3]
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Set T(t)=40, subtract 20, then solve 65e^(-0.08t)=20 by taking logarithms.
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t ≈ 14.7 minutes
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Write down the temperature of the room, according to the model. Justify your answer. [2]
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Examine the behavior of T(t) as t becomes very large.
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The room temperature is 20 °C.
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Find the rate at which the coffee is cooling after 5 minutes. [3]
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Differentiate T(t) to get T'(t) and evaluate at t=5.
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T'(5) ≈ -3.49 °C/min
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Explain what \(T'(t)\) represents in the context of this model. [1]
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Interpret the derivative in real-world terms.
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It represents the instantaneous rate of change of the coffee’s temperature (°C/min), i.e. how quickly it is cooling.
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