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.
3.
What was the initial temperature of the soup?
4.
What is the temperature of the soup after 15 minutes?

Answer 1

@jim_thompson5910 help?

Answer 2

plug in t = 0. What do you get for T?

Answer 3

212 degrees

Answer 4

correct. This is the initial temperature because it's the temperature when the time is t = 0 minutes

Answer 5

`What is the temperature of the soup after 15 minutes?`
now plug in t = 15

Answer 6

im stuck can you help? @jim_thompson5910

Answer 7

show me what you have so far

Answer 8

\[T(15)=68+144e ^{-.04(15)}\]

Answer 9

good, now you just need to use a calculator to evaluate `68 + 144*e^(-0.04*15)`

Answer 10

\[T(15)=68+144e ^{-0.6}\]

Answer 11

btw the 'e' is a constant like 'pi' is (just a different fixed number)

Answer 12

okay

Answer 13

what do i do now?

Answer 14

what is the value of `68 + 144*e^(-0.04*15)`

use a calculator. If you don't have one, use this

http://web2.0calc.com/

Answer 15

147

Answer 16

did you get a whole number? or decimal number?

Answer 17

this is what the calculator said 147.028875597539816

Answer 18

I'm getting `147.0288756` so it looks like we both got the same

Answer 19

if you round to the nearest whole degree, then yes, it's 147 degrees F

Answer 20

so it will be 147 degrees F when it has been 15 min?

Answer 21

yes

Answer 22

yay! thanks

Answer 23

assuming they want you to round to the nearest whole number

Answer 24

no problem