Question

1. a hot bowl of soup cools according to newtons law of cooling. its temperature (in °f ) at time t is given by (t(t)=68 + 144e^{-0.04t}), where t is given in minutes. part i: what was the initial temperature (at time t = 0) of the soup? (4 points) part ii: what is the temperature of the soup when t = 15 minutes? (4 points) part iii: to the nearest whole minute, at what time t is the soups temperature 125 degrees fahrenheit? (4 points)

Explanation:

Step1: Find initial temperature (t = 0)

Substitute \(t = 0\) into \(T(t)=68 + 144e^{-0.04t}\). Since \(e^0=1\), we have \(T(0)=68+144\times1\).
\[T(0)=68 + 144=212\]

Step2: Find temperature at t = 15

Substitute \(t = 15\) into \(T(t)=68 + 144e^{-0.04t}\). First, calculate the exponent: \(-0.04\times15=-0.6\). Then find \(e^{- 0.6}\approx0.5488\). So \(T(15)=68+144\times0.5488\).
\[T(15)=68 + 144\times0.5488=68+79.0272 = 147.0272\approx147\]

Step3: Find t when T(t) = 125

Set \(T(t)=125\), so \(125=68 + 144e^{-0.04t}\). First, subtract 68 from both sides: \(125 - 68=144e^{-0.04t}\), which gives \(57 = 144e^{-0.04t}\). Then divide both sides by 144: \(\frac{57}{144}=e^{-0.04t}\), so \(e^{-0.04t}=\frac{57}{144}\approx0.3958\). Take the natural - logarithm of both sides: \(\ln(e^{-0.04t})=\ln(0.3958)\). Since \(\ln(e^{-0.04t})=-0.04t\), we have \(-0.04t=\ln(0.3958)\approx - 0.925\). Then solve for \(t\): \(t=\frac{-0.925}{-0.04}=23.125\approx23\)

Answer:

Part I: 212
Part II: 147
Part III: 23