Separable Differential Equations

Question

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 205 degrees Fahrenheit when freshly poured, and 3 minutes later has cooled to 195 degrees in a room at 76 degrees, determine when the coffee reaches a temperature of 150 degrees. The coffee will reach a temperature of 150 degrees in how many minutes?

Answer 1

Let "T" be the object's temperature. Let "T_A" be the ambient temperature.

Since "T" is the object's temp, then dT/dt is the rate at which the object's temperature is changing.

We are told the dT/dt is proportional to (T - T_A) so we can say:

dT/dt = k(T - T_A) where "k" is the constant of proportionality

So dT/(T - T_A) = kdt

Integrating: ∫1/(T - T_A) dT = ∫k dt

ln(T - T_A) = kt + C₁

Since a logarithm is an exponent and the base of the logarithm is "e" then this becomes:

T - T_A = e^{kt + C1}

And since adding exponents infers multiplying terms with the same base then:

T - T_A = e^kt • e^C1

Since e^C1 is just a constant, we can call it C₂ so the equation becomes:

T - T_A = e^kt • C₂ or T - T_A = C₂(e^kt)

That gives us T as a function of time:

T(t) = T_A + C₂(e^kt)

We are given certain data: T_A = 76 and T_initial = 205 meaning at time t = 0, the temp is 205 F so:

205 = 76 + C₂(e^k(0))

205 = 76 + C₂(1)

C₂ = 129 so:

T(t) = 76 + 129(e^kt)

We are also told that after 3 minutes, T = 195 so:

195 = 76 + 129(e^k(3))

195 - 76 = 129e^3k

119/129 = e^3k

0.92248 = e^3k

ln(0.92248) = ln(e^3k)

-0.08066889 = 3k

k = -0.0268963

So T(t) = 76 + 129(e^-0.0268963t)

Now that we have the equation, let's find out when T = 150:

150 = 76 + 129(e^-0.0268963t)

150 - 76 = 129(e^-0.0268963t)

74/129 = e^-0.0268963t

ln(74/129) = ln(e^-0.0268963t)

-0.5557473 = -0.0268963t

t = 20.66 minutes