Question

a person knows that to get well - done beef, it should be brought to a temperature of ( 173^{circ}f ). he placed a sirloin tip roast with a temperature of ( 35^{circ}f ) in an oven with a temperature of ( 328^{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 ( 173^{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 ( 328^{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 2 hour 4 minutes longer.
(round to the nearest whole number as needed.)
b. the roast will never get well done.

Explanation:

Step1: Set up the exponential - decay formula for temperature difference

The formula for the temperature difference \(T(t)\) (where \(T(t)\) is the difference between the oven temperature \(T_{oven}\) and the roast temperature \(T_{roast}(t)\)) is \(T(t)=T_{0}e^{-kt}\), where \(T_{0}=T_{oven}-T_{roast}(0)\)

Given \(T_{oven} = 328^{\circ}F\), \(T_{roast}(0)=35^{\circ}F\), so \(T_{0}=328 - 35=293\)

After \(t = 4\) hours, \(T_{roast}(4)=139^{\circ}F\), and \(T(4)=328 - 139 = 189\)

Substitute into the formula \(T(t)=T_{0}e^{-kt}\): \(189 = 293e^{-4k}\)

Step2: Solve for \(k\)

Divide both sides of the equation \(189 = 293e^{-4k}\) by \(293\): \(\frac{189}{293}=e^{-4k}\)

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

\(k=-\frac{1}{4}\ln(\frac{189}{293})\approx-\frac{1}{4}\times(- 0.439)\approx0.11\)

Step3: Find the time \(t\) when \(T_{roast}(t)=173^{\circ}F\)

When \(T_{roast}(t) = 173^{\circ}F\), \(T(t)=328 - 173=155\)

Substitute into \(T(t)=T_{0}e^{-kt}\): \(155 = 293e^{-0.11t}\)

Divide both sides by \(293\): \(\frac{155}{293}=e^{-0.11t}\)

Take the natural logarithm of both sides: \(\ln(\frac{155}{293})=-0.11t\)

\(t=\frac{\ln(\frac{155}{293})}{- 0.11}=\frac{-0.619}{-0.11}\approx5.63\) hours

The additional time \(\Delta t=5.63 - 4=1.63\) hours

Convert \(0.63\) hours to minutes: \(0.63\times60 = 38\) minutes (approx)

Answer:

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