Question

a person knows that to get well - done beef, it should be brought to a temperature of ( 170^{circ}f ). he placed a sirloin tip roast with a temperature of ( 40^{circ}f ) in an oven with a temperature of ( 329^{circ}f ) and after 4 hours the temperature of the roast was ( 139^{circ}f ). how much longer must the roast be in the oven to get it well done? if the oven temperature is set at ( 170^{circ}f ), how long will it take to get the roast well done? hint: the difference between the roast temperature and the oven temperature decreases exponentially.
how much longer will it take to get the roast well done with an oven temperature of ( 329^{circ} )? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the roast must be in the oven 1 hour 42 minutes longer.
(round to the nearest whole number as needed.)
b. the roast will never get well done.
if the roast is initially placed in an oven whose temperature is set at ( 170^{circ}f ), how long will it take to get the roast well done? select the correct choice below and, if necessary, fill in the answer boxes to complete your choice.
a. the roast must be in the oven hour minutes longer.
(round to the nearest whole number as needed.)
b. the roast will never get well done.

Explanation:

Step1: Newton's Law of Heating

Newton's Law of Heating is given by the formula \(T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}\), where \(T(t)\) is the temperature of the object at time \(t\), \(T_{s}\) is the surrounding temperature, \(T_{0}\) is the initial temperature of the object, and \(k\) is a constant.

When \(T_{s} = 329^{\circ}F\), \(T_{0}=40^{\circ}F\). After \(t = 4\) hours, \(T(4)=139^{\circ}F\).
Substitute these values into the formula:
\(139=329+(40 - 329)e^{-4k}\)
\(139-329=( - 289)e^{-4k}\)
\(-190=-289e^{-4k}\)
\(e^{-4k}=\frac{190}{289}\)
Take the natural logarithm of both sides: \(-4k=\ln(\frac{190}{289})\)
\(k=-\frac{1}{4}\ln(\frac{190}{289})\approx-\frac{1}{4}(-0.457)\approx0.114\)

Step2: Find the total time \(t\) when \(T(t) = 170^{\circ}F\)

Substitute \(T(t)=170\), \(T_{s} = 329\), \(T_{0}=40\) and \(k = 0.114\) into \(T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}\)
\(170=329+(40 - 329)e^{-0.114t}\)
\(170-329=( - 289)e^{-0.114t}\)
\(-159=-289e^{-0.114t}\)
\(e^{-0.114t}=\frac{159}{289}\)
Take the natural logarithm of both sides: \(-0.114t=\ln(\frac{159}{289})\)
\(t=\frac{\ln(\frac{159}{289})}{- 0.114}\approx\frac{-0.607}{-0.114}\approx5.32\) hours

Since it has already been in the oven for \(4\) hours, the additional time is \(t-4=5.32 - 4=1.32\) hours. \(0.32\) hours is \(0.32\times60 = 19.2\approx20\) minutes. But if we use the exact value of \(k\) from \(e^{-4k}=\frac{190}{289}\), \(k=\frac{\ln(289)-\ln(190)}{4}\)

\(T(t)=170\), \(170 = 329+(40 - 329)e^{-kt}\)
\(e^{-kt}=\frac{170 - 329}{40 - 329}=\frac{-159}{-289}\)
\(t=\frac{\ln(\frac{289}{159})}{k}\), and since \(k=\frac{\ln(\frac{289}{190})}{4}\)
\(t = 4\times\frac{\ln(\frac{289}{159})}{\ln(\frac{289}{190})}\approx4\times\frac{0.607}{0.457}\approx5.32\)

The additional time is \(1.32\) hours. \(0.32\) of an hour is \(0.32\times60 = 19.2\approx20\) minutes. But if we calculate more accurately:

Let \(y = e^{-kt}\), from \(T(t)=T_{s}+(T_{0}-T_{s})y\)

For the first part with \(T_{s}=329\), \(T_{0} = 40\)

We have \(T(t)=329-289e^{-kt}\)

When \(t = 4\), \(139=329-289e^{-4k}\), \(e^{-4k}=\frac{329 - 139}{289}=\frac{190}{289}\)

When \(T(t)=170\), \(170=329-289e^{-kt}\), \(e^{-kt}=\frac{329 - 170}{289}=\frac{159}{289}\)

Since \(e^{-kt}=(e^{-4k})^{\frac{t}{4}}\), \(\frac{159}{289}=(\frac{190}{289})^{\frac{t}{4}}\)

Take the logarithm of both sides: \(\ln(\frac{159}{289})=\frac{t}{4}\ln(\frac{190}{289})\)

\(t = 4\times\frac{\ln(\frac{159}{289})}{\ln(\frac{190}{289})}\approx4\times\frac{- 0.607}{-0.457}\approx5.32\)

The additional time is \(1.32\) hours. \(1.32\) hours \(=1\) hour \(+0.32\times60\) minutes \(=1\) hour \(19.2\) minutes \(\approx1\) hour \(20\) minutes. But if we use the formula \(T(t)-T_{s}=(T_{0}-T_{s})e^{-kt}\)

For the second part, when \(T_{s}=170\), \(T(t) = 170\), \(T_{0}=40\)
\(T(t)-T_{s}=(T_{0}-T_{s})e^{-kt}\)
\(170 - 170=(40 - 170)e^{-kt}\)
\(0=- 130e^{-kt}\)
Since \(e^{-kt}>0\) for all real \(t\), the equation \(0=-130e^{-kt}\) has no solution.

Answer:

For the oven temperature of \(329^{\circ}F\): A. The roast must be in the oven \(1\) hour \(20\) minutes longer.
For the oven temperature of \(170^{\circ}F\): B. The roast will never get well - done.