This exercise uses Newton's Law of Cooling. A hot bowl of soup is served at a d

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This exercise uses Newton's Law of Cooling.

A hot bowl of soup is served at a dinner party. It starts to cool according to Newton's Law of Cooling so that its temperature at time t is given by
T(t) = 51 + 132e−0.05t
where t is measured in minutes and T is measured in °F.
(a) What is the initial temperature of the soup?
 °F

(b) What is the temperature after 10 min? (Round your answer to one decimal place.)
 °F

(c) After how long will the temperature be 100°F? (Round your answer to the nearest whole number.)

 min

Jul 15th, 2015

Thank you for the opportunity to help you with your question!

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where t is measured in minutes and T is measured in °F.


a) To find the initial temperature put time  t = 0


So, initial temperature = 183 °F .


b) To find the temperature after 10 minutes put time t = 10


So, temperature after 10 minutes = 131.1 °F .


c) To find after how long will the temperature be 100°F put T(t) = 100


So, it will take 20 minutes for the temperature to become 100 °F .

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Please let me know if you need any clarification. I'm always happy to answer your questions.
Jul 15th, 2015

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