Question

when you remove soup from a crock pot, its temperature is 200°f. the room temperature is 72°f, and the cooling rate of the soup is r = 0.054. use newtons law of cooling to find how long it will take the soup to cool to a serving temperature of 100°f. round your answer to the nearest minute.
it will take about
minutes.

Explanation:

Step1: Recall Newton's Law of Cooling formula

Newton's Law of Cooling is given by \( T(t)=T_{a}+(T_{0}-T_{a})e^{-rt}\), where \( T(t)\) is the temperature at time \( t\), \( T_{a}\) is the ambient (room) temperature, \( T_{0}\) is the initial temperature, \( r\) is the cooling rate.
Here, \( T_{0} = 200^{\circ}F\), \( T_{a}=72^{\circ}F\), \( r = 0.054\), and \( T(t)=100^{\circ}F\).
Substitute these values into the formula: \( 100=72+(200 - 72)e^{-0.054t}\).

Step2: Simplify the equation

First, simplify \( 100=72+(200 - 72)e^{-0.054t}\) to \( 100-72 = 128e^{-0.054t}\).
So, \( 28=128e^{-0.054t}\).
Then, \( e^{-0.054t}=\frac{28}{128}=\frac{7}{32}\).

Step3: Take the natural logarithm of both sides

Take the natural - logarithm of both sides: \(\ln(e^{-0.054t})=\ln(\frac{7}{32})\).
Using the property \(\ln(e^{x})=x\), we get \(- 0.054t=\ln(\frac{7}{32})\).
Since \(\ln(\frac{7}{32})=\ln(7)-\ln(32)\approx1.9459 - 3.4657=-1.5198\).

Step4: Solve for \( t\)

We have \(-0.054t=-1.5198\).
Then \( t=\frac{-1.5198}{-0.054}\approx28.14\).

Answer:

\(28\)