Question

You place a cup of 200ºF coffee on a table in a room that is 67ºF, 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)=T(subscriptA)+(T[subscripto]−T[subscriptA])e^(−kt)

A. 40 minutes
B. 15 minutes
C. 1 hour
D. 35 minutes

Answer 1

ah back to newton's law yes?

Answer 2

wanna do it the easy way?

Answer 3

T(t) = final temperature, T(a) = ambient temp (that is, the temperature of the room), T(to) = original temperature, k = constant you need to solve for, t = time

Answer 4

just plug the the variables from the first sentence into the equation, solve for k, and then plug the variables with the k value you calculated, only this time use 180 instead of 195 as the final temperature, and solve for t this time

Answer 5

you understand all that or you want to do it step by step or you want the answer?

Answer 6

the answer would be nice lol

Answer 7

give me as second i am going to do it the easy way, then give you the answer

Answer 8

i get 42 so i guess 40 is closest

Answer 9

let me check calculator again

Answer 10

yup. that is what i get. you want easy method?

Answer 11

thanks, and sure

Answer 12

ok first off we need to work with the differences in the temp, not the actual temp. don't let this throw you off
200- 67 = 133
195-67 = 128
185 - 67 = 113

Answer 13

those are the numbers we will work with on the right.

Answer 14

decreased from 133 to 128 in ten minutes formula is
\[133(\frac{128}{133})^{\frac{t}{10}}\]

Answer 15

you see how i put the numbers in. the ten is from the ten minutes, the initial values is 133 and then end value is 128

Answer 16

then solve
\[113=133(\frac{128}{133})^{\frac{t}{10}}\] for t and if that is hard just do it here
http://www.wolframalpha.com/input/?i=113%3D133%28128%2F133%29^%28x%2F10%29

Answer 17

cool, thanks