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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.

AWhat was the initial temperature of the soup?
BWhat is the temperature of the soup after 15 minutes?
CHow long after serving is the soup 125 degrees F?

Verified Answer :

antonsandiego

The initial temperature is get when t=0 namely
So, we have \(68+144e0=68+144\) which is a point of boiling water in 212 degrees
Then calculate 15 minutes using the first mentioned formula 
You should have  T(15)=68+144e^(-.04*15)=147.03 degrees

Answer:   212°F

147°F

23.2 min

1) Initial temperature for t=0

T(0)=68+144e^(0)=68+144=212°F

2) After 15 min

T(15)=68+144e^(-0.04*15)=68+144e^(-0.6)=147°F

3) 125°F -> t=?

125=68+144e^(-0.04t)

144e^(-0.04t)=125-68=57

e^(-0.04t)=57/144=0.4 you can apply ln on both sides:

-0.04t=ln(0.4) solving you get t=23.3 min

Explanation:

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