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Question

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.)

(a) What is the temperature of the drink after 60 minutes?

(b) When will its temperature be 16°C?

Answer 1

This problem is a typical application of differential equation under the name of Newton's law of cooling. Which states that the rate of change of temperature is directly proportional to the difference in temperature between two environments(systems). Mathematical represented as

dT/dt = kΔT = k(T - 20) ------------ (1) [Where T: Temperature, t: time, k: proportionality constant]

This differential equation has solution of the form

T(t) = Ae^kt + 20 ------------ (2)

Solve for A by applying the initial condition given i.e

T(t=0) = 5 = Ae^k*0 +20 => -15 = A

Thus equation (2) becomes

T(t) = 20 - 15e^kt

Using the condition T(t=25mins) = 10°C, we find the constant k as

10 = 20 - 15e^kt => k = ln(2/3)/25 ≈ -0.016

a) After 60 mins, you have

T(t=60mins) = 20 - 15e^-0.016*60 ≈ 14.33°C

b) Set T(t=?) = 16°C, then from equation (2)

16 = 20 - 15e^-0.016t [Solving for t gives]

(16 -20) = - 15e^{-0.016 t} => t = ln(4/15)/(-0.016) ≈ 81.50 minutes.

Answer 2

Let T_d = temperature of the drink

T_r = temperature of the room

T_do = temperature of the drink at time = 0

The governing equation will be T_d = (T_do - T_r)e^kt + 20 where k = the proportionality constant

at T_d(0) = (5 - 20) + 20 = 5 degrees (OK)

at T_d(25) = (5 - 20)e^25k + 20 = 10 Solving for k, we find that k = -.016219

So the equation becomes

T_d(t) = -15e^(-.016219t) + 20

If we plug in for 60 minutes, we find that T_d = 14.33 degrees

If you plug in 16 degrees for T_d, you find that t = 81.5 minutes.