A can of soda at room temperature 70◦F is placed in a refrigerator that maintains a constant temperature of 40◦F. After 1 hour in the refrigerator, the temperature of the soda is 58◦F. At what time will the soda’s temperature be 41◦F?

Question

Accepted Answer

Let T(t ) denote the temperature of the soda at time t in degrees F; note that [latex]T_{0}[/latex] = 70. Since the surrounding temperature is 40, T satisfies the initial-value problem

[latex]T´=−k(T −40), T(0) = 70[/latex]

and therefore by (2.4.12) T has the form

[latex]T(t ) = (T_{0} −T_{m})e^{−kt} +T_{m}[/latex] (2.4.12)

[latex]T(t ) =30e^{−kt} +40[/latex]

In particular, note that the temperature is decreasing exponentially as time increases and tending towards 40◦F, the temperature of the refrigerator, as t →∞.

To determine the constant k, we use the additional given information that T(1) = 58, and therefore

[latex]58 =30e^{−kt} +40[/latex]

It follows that [latex]e^{−k}[/latex] = 3/5, and thus k = ln(5/3). To now answer the original question, we solve the equation

[latex]41 = 30e^{−ln(5/3)t} +40[/latex]

and find that t = ln(30)/ln(5/3) ≈ 6.658 h.

Question 2.4.3: A can of soda at room temperature 70◦F is placed in a refrig...

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