A roasted turkey is taken from an oven when its temperature has reached 185 Fahrenheit and is...

Question:

A roasted turkey is taken from an oven when its temperature has reached 185 Fahrenheit and is placed on a table in a room where the temperature is 75 Fahrenheit.

(a) If the temperature of the turkey is 157 Fahrenheit after half an hour, what is its temperature after 45 minutes?

(b) When will the turkey cool to 100 Fahrenheit? (In hours)

Newton's Cooling Law

When an object and the environment have different temperatures, the temperature of the object changes over time until its temperature is the same as the environment. This is because the environment is large enough that it can act as a heat bath, where is absorbs or gives off heat without changing its temperature. The change in the temperature of the object over time is described by Newton's cooling law equation.

Answer and Explanation:


The temperature of the turkey is given by Newton's cooling law equation

{eq}\displaystyle T(t) = T_s + (T_0 - T_s) e^{-kt} {/eq}

where {eq}T_s {/eq} is the temperature of the room, {eq}T_0 {/eq} is the initial temperature of the turkey and k is the cooling constant to be determined.


We have the following given:

{eq}\displaystyle \begin{align*} \text{Initial temperature of the turkey: } T_0 &= 185^{\circ}\ F \\ \text{Temperature of the room: } T_s &= 75^{\circ}\ F\\ \text{Temperature of the turkey after 30 minutes: } T(30) &= 157^{\circ}\ F \end{align*} {/eq}


To solve this problem, we need to solve for the explicit form of Newton's cooling law equation for this problem.


We start with plugging the given into the general equation:

{eq}\displaystyle \begin{align*} T(t) &= T_s + (T_0 - T_s) e^{-kt} \\ &= 75^{\circ}\ F + ( 185^{\circ}\ F - 75^{\circ}\ F) e^{-kt} \\ T(t) &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-kt} \end{align*} {/eq}


Now we solve for k using the fact that at {eq}t = 30\ min,\ T(30\ min) = 157^{\circ}\ F {/eq}.


{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-kt},\ T(30\ min)= 157^{\circ}\ F \\ T(30\ min) &=75^{\circ}\ F + ( 110^{\circ}\ F) e^{-k(30\ min)} \\ 157^{\circ}\ F &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-k(30\ min)} \\ 157^{\circ}\ F- 75^{\circ}\ F &= ( 110^{\circ}\ F) e^{-k(30\ min)} \\ \frac{82^{\circ}\ F}{ ( 110^{\circ}\ F)}&= e^{-k(30\ min)} \\ \end{align*} {/eq}


We can solve for k by taking the natural logarithm of both sides of the equation.

{eq}\displaystyle \begin{align*} \frac{82}{110}&= e^{-k(30\ min)} \\ \ln \bigg[ \frac{82}{110}&= e^{-k(30\ min)} \bigg] \\ \ln \frac{82}{110} &=-k(30\ min) \\ k &= \boxed{-\frac{1}{30}\ln \frac{82}{110}\ min^{-1} \text{ or } 0.009792\ min^{-1} } \end{align*} {/eq}


Therefore, we can write the explicit form of the Newton's cooling law equation for this problem as

{eq}\displaystyle T(t) = 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})t} {/eq}.


(a.) To solve for the temperature of the turkey after 45 minutes, we plug {eq}t = 45\ min {/eq} into {eq}T(t) {/eq}.


{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})t},\ t = 45\ min\\ &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})(45\ min)}\\ &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})(45\ min)}\\ &= 75^{\circ}\ F +70.80^{\circ}\ F \\ T(45\ min) &= \boxed{ 145.80^{\circ}\ F} \end{align*} {/eq}


Therefore, at {eq}t = 45\ min {/eq} the temperature of the turkey is {eq}\boxed{ 145.80^{\circ}\ F}. {/eq}


(b) To solve for the time it takes for the turkey to cool to {eq}100^{\circ}\ F {/eq}, we plug this into {eq}T(t) = 100^{\circ}\ F {/eq} and solve for t.


{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})t},\ T(t) = 100^{\circ}\ F\\ 100^{\circ}\ F &= 75^{\circ}\ F + ( 110^{\circ}\ F) e^{-(0.009792\ min^{-1})t}\\ 100^{\circ}\ F - 75^{\circ}\ F &= 110^{\circ}\ F e^{-(0.009792\ min^{-1})t}\\ \frac{25^{\circ}\ F}{110^{\circ}\ F} &= e^{-(0.009792\ min^{-1})t} \end{align*} {/eq}


Taking the natural logarithm of both sides to solve for t.


{eq}\displaystyle \begin{align*} \ln \bigg[ \frac{25^{\circ}\ F}{110^{\circ}\ F} &= e^{-(0.009792\ min^{-1})t}\bigg] \\ \ln \frac{25}{110} &=-(0.009792\ min^{-1})t\\ t &= \frac{1}{-0.009792\ min^{-1}}\ln \frac{25}{110} \\ t &= \boxed{ 151.3\ min} \end{align*} {/eq}


Therefore, it will take around {eq}\boxed{ 151.3\ min} {/eq} for the turkey to cool to {eq}100^{\circ}\ F {/eq}.


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