##### 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

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$T(t) = 51 + 132e^{-0.05t}$

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

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

$\\ T(0) = 51 + 132e^{-0.05\times0}\\ \\ T(0) = 51 + 132e^{0}\\ \\ T(0) = 51 + 132\times1\\ \\ T(0) = 51 + 132\\ \\ T(0) = 183\hspace{5}^{\circ}F$

So, initial temperature = 183 °F .

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

$\\ T(10) = 51 + 132e^{-0.05\times10}\\ \\ T(10) = 51 + 132e^{-0.5}\\ \\ T(10) = 131.1\hspace{5}^{\circ}F$

So, temperature after 10 minutes = 131.1 °F .

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

$\\ 100 = 51 + 132e^{-0.05t}\\ \\ 100 - 51 = 132e^{-0.05t}\\ \\ 49 = 132e^{-0.05t}\\ \\ 132e^{-0.05t}=49\\ \\ e^{-0.05t}=\frac{49}{132}\\ \\ Take\hspace{5}natural\hspace{5}logarithm\hspace{5}on\hspace{5}both\hspace{5}sides\\ \\ ln(e^{-0.05t})=ln\bigg(\frac{49}{132}\bigg)\\ \\ -0.05t\times ln(e)=ln\bigg(\frac{49}{132}\bigg)\\ \\ -0.05t\times1=ln\bigg(\frac{49}{132}\bigg)\\ \\ -0.05t=ln\bigg(\frac{49}{132}\bigg)\\ \\ t=\frac{ln\bigg(\frac{49}{132}\bigg)}{-0.05}\\ \\ t=19.82\hspace{5}min\\ \\ t=20\hspace{5}min(rounded \hspace{5}to\hspace{5} the\hspace{5} nearest\hspace{5} whole\hspace{5} number)\\$

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

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Jul 15th, 2015

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Jul 15th, 2015
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Jul 15th, 2015
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