please help math

Question

A hot coffee just brewed(180F) from the coffee machine is placed in a room where the temperature is 65F.

After 20 minutes, the hot coffee has cooled to 140F. Find the temperature after one hour 06 minutes.

please be very detailed. Thank you

Answer 1

Let T = temperature of coffee t minutes after pouring

By Newton's Law of Cooling, dT/dt = k(T - 65)

[1 / (T - 65)]dT = kdt

∫[1 / (T - 65)]dT = ∫kdt

ln(T - 65) = kt + C

T - 65 = e^kt+C = e^Ce^kt

Relabel e^C as C to get T = Ce^kt + 65

Since T = 180 when t = 0, C = 180 - 65 = 115.

So, T = 115e^kt + 65

Plug in T = 140 and t = 20 to solve for k.

Then, evaluate T when t = 66

Answer 2

This requires solution of a differential equation unless you have been given a formula for cooling.

The equation is dT/dt=-k(T-65) which says that the rate of cooling is proportional to the temperature difference between the object and the environment. Note t=time in hours!

This is a variables separable equation the general solution of which is

ln(T-65)=-kt+K => T-65=Ce^-kt

The initial condition gives C=180-65=115.

The proportionality constant k is determined by setting t=1/3 and solving the equation of temperature for k.

Then you have the final equation of cooling and you put t=1.1 to find the answer required.

T-65