Question

a person knows that to get well - done beef, it should be brought to a temperature of 173°f. he placed a sirloin tip roast with a temperature of 35°f in an oven with a temperature of 328°f and after 4 hours the temperature of the roast was 139°f. how much longer must the roast be in the oven to get it well done? if the oven temperature is set at 173°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 328°? 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 Formula

Newton's Law of Heating is given by \(T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}\), 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 the cooling/heating constant.

Here, \(T_{s} = 328^{\circ}F\), \(T_{0}=35^{\circ}F\). When \(t = 4\) hours, \(T(4)=139^{\circ}F\).

Substitute into the formula: \(139=328+(35 - 328)e^{-4k}\)

First, simplify the equation:

\(139-328=(35 - 328)e^{-4k}\)

\(- 189=-293e^{-4k}\)

\(e^{-4k}=\frac{189}{293}\)

Take the natural logarithm of both sides: \(\ln(e^{-4k})=\ln(\frac{189}{293})\)

Since \(\ln(e^{x})=x\), we have \(-4k=\ln(\frac{189}{293})\)

\(k=-\frac{1}{4}\ln(\frac{189}{293})\approx-\frac{1}{4}(- 0.419)\approx0.105\)

Step2: Find the time \(t\) when \(T(t) = 173^{\circ}F\)

Substitute \(T(t)=173\), \(T_{s} = 328\), \(T_{0}=35\) and \(k = 0.105\) into \(T(t)=T_{s}+(T_{0}-T_{s})e^{-kt}\)

\(173=328+(35 - 328)e^{-0.105t}\)

\(173-328=(35 - 328)e^{-0.105t}\)

\(-155=-293e^{-0.105t}\)

\(e^{-0.105t}=\frac{155}{293}\)

Take the natural logarithm of both sides: \(\ln(e^{-0.105t})=\ln(\frac{155}{293})\)

Since \(\ln(e^{x})=x\), we have \(-0.105t=\ln(\frac{155}{293})\)

\(t=\frac{\ln(\frac{155}{293})}{- 0.105}\)

\(\ln(\frac{155}{293})\approx\ln(0.529)\approx - 0.637\)

\(t=\frac{-0.637}{-0.105}\approx6.07\) hours

The additional time is \(t-4\approx6.07 - 4=2.07\) hours

\(0.07\times60 = 4.2\) minutes

Answer:

A. The roast must be in the oven \(2\) hour \(4\) minutes longer.