Question

this exercise uses newtons law of cooling.
a hot bowl of soup is served at a dinner party. it starts to cool according to newtons law of cooling so that its temperature at time ( t ) is given by
t(t)=57 + 149 e^{-0.05 t}
where ( t ) is measured in minutes and ( t ) is measured in ( { }^{circ} mathrm{f} ).
(a) what is the initial temperature (in ( { }^{circ} mathrm{f} )) of the soup?
( { }^{circ} mathrm{f} )
(b) what is the temperature (in ( { }^{circ} mathrm{f} )) after 10 min? (round your answer to one decimal place.)
( { }^{circ} mathrm{f} )
(c) after how long (in min) will the temperature be ( 100^{circ} mathrm{f} ) ? (round your answer to the nearest whole number.)
min

Explanation:

Step1: Find the initial temperature (a)

The initial temperature is when \(t = 0\). Substitute \(t=0\) into \(T(t)=57 + 149e^{-0.05t}\).

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Step2: Find the temperature after 10 minutes (b)

Substitute \(t = 10\) into \(T(t)=57 + 149e^{-0.05t}\).

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Step3: Find the time when \(T(t)=100\) (c)

Set \(T(t)=100\), so \(100=57 + 149e^{-0.05t}\).
First, subtract 57 from both sides: \(100 - 57=149e^{-0.05t}\), \(43 = 149e^{-0.05t}\).
Then, divide both sides by 149: \(\frac{43}{149}=e^{-0.05t}\).
Take the natural logarithm of both sides: \(\ln(\frac{43}{149})=- 0.05t\).

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Answer:

a. \(206\)
b. \(147.4\)
c. \(25\)