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A hot bowl of soup cools according to Newton’s law of cooling. Its temperature (in ∞F) at time t is given by T(t) = 76+105e^-0.125t, where t is given in minutes. 

 

What was the initial temperature of the soup?

 

 

 

What is the temperature of the soup after 30 minutes?

 

Thank you~~~

 Dec 18, 2020
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Initial temperature of the soup is when t = 0

76 + 105 e-0.125 (0) =  76 + 105 = 181 degrees F

 

After 30 miutes:

76 + 105 e-0.125(30= 76 + 2.47 =78.47 degrees F    

 Dec 18, 2020

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