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 the function
t(t)=68+144e^-0.04t

, where t is expressed in minutes. At what time will the soup reach 200oF?
A. 137.39 minutes
B. 2.18 minutes
C. 16.39 minutes
D. 1.02 minutes

Answer 1

Set the equation equal to 200 then subtract 200 from each side and solve the quadratic equation. I'm going to assume that you need the quadratic formula to continue.

Answer 2

problem is idk how to solve it

Answer 3

ok.... you should set t(t) = 200

so

\[200 = 68 + 144e^{-0.04t}\]


set 1 subtract 68 from both sides.... what do you get..?

Answer 4

-132+144e^-0.04t

Answer 5

nearly

\[132 = 144e^{-0.04t}\]

you are solving for t like any other equation, so you want t on one side and the value on the other side...

next step divide both sides by 144

Answer 6

.916=e^-0.04t

Answer 7

ok.... so do you know log laws...?

particularly

\[\ln{e^a} = a\]

Answer 8

nope

Answer 9

i have my calculator though

Answer 10

ok.... well you'll have to trust me... do you have a calculator with an ln function

Answer 11

yes

Answer 12

pk this is what you need to find

ln(132/144) it will be a negative answer...

Answer 13

-.087011377

Answer 14

i think you're missing a zero after the decimal point...

but thats ok

by taking the ln on both sides the exponential disappears...

so

ln(132/144) = - 0.04t

or

-0.087011 = -0.04t

last step is to divide both sides by -0.04

Answer 15

2.175284425

Answer 16

great... that is how many minutes it takes the soup to cool from 212 to 200 degrees...

when t = 0 the temperature of the soup is

\[t(t) = 68 + 144e^{-0.04 \times 0} = 68 + 144 = 212\]

Answer 17

so 2.1752... minutes later the soup is 200 degree

Answer 18

hope it makes sense...