When a cold drink is taken from a refrigerator, its temperature is 5ºC. After 25 minutes in a 20ºC room its temperature has increased to 10 ºC.

Question

When a cold drink is taken from a refrigerator, its temperature is 5ºC. After 25 minutes in a 20ºC room its temperature has

increased to 10 ºC.

(a) What is the temperature of the drink after 50 minutes?
(b) When will its temperature be 15ºC?

 

 

ANSWER: (a) 13.3ºC, (b) 67.74 min

#15, p242

Answer 1

The heating (T) of an object cooler than its surrounding medium (20°C room) with respect to time (t) is described by dT/dt = k(20 − T) or dT/dt + kT = 20k.

Obtain the Integrating Factor for this Linear Differential Equation by writing I(t) = e^∫kdt or e^kt.

Next, e^kt[dT/dt + kT = 20k] gives e^ktdT/dt + ke^ktT = 20ke^kt which amounts to d(Te^kt)/dt = 20ke^kt.

Then, by integration, Te^kt = 20e^kt + C.

Multiply through this last equation by e^-kt to yield T or T(t) = 20 +Ce^-kt .

T is 5° at t = 0 minutes, so write T(0) = 5 = 20 + Ce^(-k•0) or C = -15.

Obtain k by T(25) = 20 + -15e^-25k = 10 or e^-25k = -10/-15; then -25k equals ln (2/3) or k = 0.01621860432.

Then for T = 50°C, write T(50) = 20 + -15e^{(-50•0.01621860432)} or 13.333333333°C equivalent to 13.3°C.

For time at T = 15°C, write 15 = 20 − 15e^{-0.01621860432t} which simplifies to -5/-15 = 1/3 = e^{-0.01621860432t} or

t = ln(1/3)/-0.01621860432 or 67.7377823 minutes equivalent to 67.74 minutes.