Question

Please help!
You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling:
T(t)=TA+(To−TA)e−kt


A. 15 minutes
B. 25 minutes
C. 1 hour
D. 45 minutes

Answer 1

plug in your numbers and solve for k. TA is the ambient temp, 72 and T0 is 205 the initial temp
\(T(t) = 72+(205-72)e^{-kt}\)

\(T(t) = 72+(178)e^{-kt}\)

Since we know T = 195 at t = 10, you can solve for k

\(195 = 72+(178)e^{-10k}\)

Answer 2

Once you have k, plug it into the equation for T(t) and solve for t when T = 180

Answer 3

Is k 0.037?

Answer 4

yes

Answer 5

then you can do
\[180=72+178e^{-0.03696t}\]
BTW, the more digits you keep for k, the more accurate your t will be.

Answer 6

Is it a?

Answer 7

I got 13.5187 :/

Answer 8

yeah that's what I got. I'm wondering if you're supposed to convert the units to celsius or seconds

Answer 9

i have no idea. it doesn't say