Question

when you remove soup from a crock pot, its temperature is 180°f. the room temperature is 74°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 95°f. round your answer to the nearest minute.
it will take about (square) minutes.

Explanation:

Step1: Write Newton's Law of Cooling formula

Newton's Law of Cooling formula is \( T(t)=T_{a}+(T_{0} - T_{a})e^{-rt}\), 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, \(T_{0}=180\), \(T_{a} = 74\), \(r=0.054\), and \(T(t)=95\).
Substitute the values into the formula: \(95=74+(180 - 74)e^{-0.054t}\).

Step2: Simplify the equation

First, simplify \(95 - 74=(180 - 74)e^{-0.054t}\).
\(21 = 106e^{-0.054t}\). Then, \(\frac{21}{106}=e^{-0.054t}\).

Step3: Take the natural logarithm of both sides

Take \(\ln\) of both sides: \(\ln(\frac{21}{106})=\ln(e^{-0.054t})\).
Using the property \(\ln(e^{x})=x\), we get \(\ln(\frac{21}{106})=- 0.054t\).

Step4: Solve for \(t\)

We know that \(\ln(\frac{21}{106})=\ln(21)-\ln(106)\approx3.0445 - 4.6639=-1.6194\).
Then \(t=\frac{-1.6194}{- 0.054}\approx29.99\approx30\).

Answer:

\(30\)