Question

A cup of coffee at 183 degrees is poured into a mug and left in a room at 73 degrees. After 8 minutes, the coffee is 136 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt=k(T−73).
What is the temperature of the coffee after 18 minutes?

Answer 1

work with the difference in the temperatures.
\(183-73=110\) and \(136-73=63\) so your model for the temperature difference it
\[T=(\frac{63}{110})^{\frac{t}{8}}\]

Answer 2

replace \(t\) by 18 to get the difference in the temperature after 18 minutes, then add 73 to get the actual temperature

Answer 3

you can also use
\[T=73+110e^{kt}\] but that takes way more work because you have to solve an exponential equation using logs to find \(k\) . once you truncate \(k\) this method will be less accurate, so don't bother

Answer 4

Yeah i think i'm supposed to use the second method to get any marks.
Just i got stuck, since we don't know the value for dT/dt to find out the value for k.

Answer 5

ok again work with the difference in the temperatures

the model for the difference starting at time 0 will be \(110e^{kt}\) and you solve for \(k\) by replacing \(t\) by 8 and the value by 63 and solving
\[63=110e^{8k}\] for k

Answer 6

you get
\[\frac{63}{110}=e^{8k}\]
\[8k=\ln(\frac{63}{110}\]
\[k=\frac{1}{8}\times \ln(\frac{63}{110})\] and then a calculator

Answer 7

one you have that \(k\) your model for the difference in the temp will be
\[110e^{kt}\] or if you want the actual temp it is
\[73+100e^{kt}\] replace \(t\) by 18 to get the answer

Answer 8

here is k
http://www.wolframalpha.com/input/?i=ln%2863%2F110%29%2F8

Answer 9

ohhh. okay, thanks so much! :)

Answer 10

..i got the answer to be 111.896 degrees, but it turned out to be wrong. :/

Answer 11

nvm. i put in 100 instead of 110. ><
thanks again though ^^