Question

A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 65°F room. After fifteen minutes, the internal temperature of the soup was 95°F.
Use Newton's Law of Cooling to write a formula
T(t)
that models this situation, where T is the temperature of the soup in degrees Fahrenheit and t is time in minutes.

Answer 1

86.34 minutes was used   to cool the soup to 80 °F

How can we calculate this using Newton's Law of Cooling?

Newton's Law of Cooling can be computed as :

$T=T1+(T0- T1) e^{kt}$

The initial temperature  is been represent by T0 which is  ( 100 F)

The temperature of surrounding  is been given as  (69 F)

The  final temperature  is been given  as  T

t = time

we were told that the temperature of the soup move to  95 F, using  15 minutes.

Hence,  T = 95 F

t = 15 min

Then we can input the values into the above equation as :

$95= 69 + (100-69)e^{15k}$

if we simplify this , we have

$26= 31e^{15k}$

Then $e^{15k} =0.8387$

Then we can find the value of k as :

k= -0.012

Then we have the new equation as

$T= 69 + (100-69)e^{-0.012t}$

$T= 69 + 31e^{15k}$

Then at T=80F

we have,

$80= 69 + 31e^{15k}$

t=86.34

Then if we simplify , we have t=86.34 which implies that  86.34 minutes was used   to cool the soup to 80 °F

Learn more about Newton's Law of Cooling at:

brainly.com/question/17051432

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