Question

A hot bowl of soup cools according to Newton's law of cooling. Its temperature (in degrees Fahrenheit) at time t is given by T(t)=68+144e^(-.04t), where t is given in minutes.

a. What was the initial temperature of the soup?
b. What is the temperature of the soup after 15 minutes?
c. How long after serving is the soup 125 degrees F?

Answer 1

the initial temp is what you get when \(t=0\) namely \[68+144e^0=68+144\]

Answer 2

that is 212 degrees, which is the boiling point of water

Answer 3

to find the temp after 15 minutes, replace \(t\) by 15 and see what you get
use a calculator

Answer 4

is this correct? T(15)=68+144e^(-.04*15)=147.03 degrees

Answer 5

http://www.wolframalpha.com/input/?i=68%2B144e^%28-.04*15%29
that is what i get, yes

Answer 6

for the last one , solve
\[125=68+144e^{-.04t}\]for \(t\)
do you know how to do that?

Answer 7

good :)

Answer 8

for c. is this correct?


125=68+144e^(-0.4t)
(125-68)/144=0.395833333=e^(-.04t)
-0.926762032=.04t
t=23.169
After 23 minutes the soup is 125°F

Answer 9

@satellite73

Answer 10

let me check
method looks right
did you take the log at step 3?

Answer 11

yes, it is right.
looks like you got this under control right?

Answer 12

thanks!! just double checking :)

Answer 13

yw