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 189 degrees in a room at 80 degrees, determine when the coffee reaches a temperature of 144 degrees.
The coffee will reach a temperature of 144 degrees in __________ minutes.

Answer 1

easier than it looks but you have to remember to work with the temperature differences, not the temperatures

Answer 2

\[\frac{ dT }{ dt }=k(T_o - T_s) \text{ is this it?}\]

Answer 3

\(205-80=125, 189-80=109\)
you can use
\[205\times \left(\frac{109}{125}\right)^{\frac{t}{3}}\]

Answer 4

and since \(144-80=64\) your job it to solve
\[64=205\times \left(\frac{109}{125}\right)^{\frac{t}{3}}\] for \(t\)

Answer 5

ahh damn that is wrong
it is not 205 is it 125 sorry \[64=125\times \left(\frac{109}{125}\right)^{\frac{t}{3}}\]

Answer 6

T(t) = S + e^(-kt+C)
T(t) = S +(To-S)*e(-kt) where To = initial temperature at t = 0
T(t) = 80+(205-80)*e^(-kt)
T(t) = 80+125e^(-kt)
189=80+125*e^-3k
109=125*e^(-3k)

and from here is where i'm getting stuck

Answer 7

wow that is a lot of work

Answer 8

ok you can solve it this way too, takes much longer
you have to find \(k\)

Answer 9

divide by \(125\) then take the log, then divide by \(-3\) to find \(k\)

Answer 10

\[109=125e^{-3k}\]
\[\frac{109}{125}=e^{-3k}\]
\[\ln(\frac{109}{125})=-3k\]
\[k=-\frac{\ln(\frac{109}{125})}{3}\]

Answer 11

i'm ending up with t=14.6818 does that sound right to you?