As I recall, our model is
T(t) = 205 + (72-205)e^-kt
So, solve for k in T(10) = 195
Then use that k to solve for t in T(t) = 180
What do you get?
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:
1 hour
25 minutes
15 minutes
45 minutes
2 answers
oh okay i got 25 minutes, is that right?
1 answer