# A roast turkey is taken from an oven when its temperature has reached 180^o F and is placed on a ...

## Question:

A roast turkey is taken from an oven when its temperature has reached {eq}180^o F {/eq} and is placed on a table in a room where the temperature is {eq}75^o F {/eq}. (Round your answer to the nearest whole number.)

a. If the temperature of the turkey is {eq}150^o F {/eq} after half hour, what is the temperature after {eq}60 {/eq} minutes?

b. When will the turkey have cooled to {eq}95^o {/eq}?

## Newton's Cooling Law

Newton's cooling law is an exponential function that models the temperature of an object over time when placed in an environment with constant temperature. It is of the form

{eq}T(t) = T_s + (T_i- T_s)e^{-kt} {/eq}

where {eq}T_s {/eq} is the temperature of the surroundings, {eq}T_i {/eq} is the initial temperature of the object and k is the cooling constant. It can be seen that the change in temperature of an object is related to the difference between the initial temperature of the object and the temperature of the surroundings.

The temperature of the roast turkey can be modeled using Newton's cooling law equation,

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

From the given we have the following:

{eq}\begin{align*} T_s &= 75^{\circ}\ F \\ T_i &= 180^{\circ}\ F \\ T(30\ min) &= 150^{\circ}\ F. \\ \end{align*} {/eq}

We can write the partial equation for the Newton's cooling law equation for this problem as {eq}\displaystyle T(t) = 75^{\circ}\ F + (180^{\circ}\ F - 75^{\circ}\ F)e^{-kt} \\ \displaystyle T(t) = 75^{\circ}\ F + 105^{\circ}\ Fe^{-kt} . {/eq}

To solve this problem, we need to come up with the explicit form of the Newton's cooling law equation. This means we need to solve for k.

To solve for k, we use {eq}T(30\ min) = 150^{\circ}\ F\ \&\ t = 30\ min {/eq} and plug them into the partial equation.

{eq}\begin{align*} \displaystyle T(t) &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-kt},\ T(30\ min) = 150^{\circ}\ F,\ t = 30\ min \\ 150^{\circ}\ F &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-k(30\ min)}\\ 150^{\circ}\ F - 75^{\circ}\ F &= 105^{\circ}\ Fe^{-k(30\ min)}\\ 75^{\circ}\ F &= 105^{\circ}\ Fe^{-k(30\ min)}\\ e^{-k(30\ min)} &= \frac{75}{105} \end{align*} {/eq}

Taking the natural logarithm of both sides.

{eq}\begin{align*} \displaystyle \ln \bigg[ e^{-k(30\ min)} &= \frac{75}{105} \bigg] \\ -k(30\ min) &= \ln \frac{75}{105} \\ k &= -\frac{1}{30} \ln \frac{75}{105} \\ k &\approx 0.0112\ min^{-1} \end{align*} {/eq}

Therefore, the explicit form of Newton's cooling law equation for this problem is {eq}\displaystyle \boxed{ T(t) = 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})t} } {/eq}.

(a.) To solve for the temperature after 60 minutes, we plug {eq}t = 60\ min {/eq} into {eq}T(t) {/eq}.

{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})t} \\ T(60\ min) &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})(60\ min)} \\ &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})(60\ min)} \\ &= 75^{\circ}\ F + 54^{\circ}\ F \\ T(60\ min) &= \boxed{ 129^{\circ}\ F} \end{align*} {/eq}

The temperature of the roast turkey after 60 minutes is {eq}\boxed{ 129^{\circ}\ F} {/eq}.

(b.) To solve for the time required for the turkey to cool to {eq}95^{\circ}\ F {/eq}, we plug {eq}T(t) = 95^{\circ}\ F {/eq} and solve for t. The steps used in solving for t is similar to the steps in solving for k.

{eq}\begin{align*} \displaystyle T(t) &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})t},\ T(t) = 95^{\circ}\ F\\ 95^{\circ}\ F &= 75^{\circ}\ F + 105^{\circ}\ Fe^{-(0.0112\ min^{-1})t}\\ 95^{\circ}\ F - 75^{\circ}\ F &= 105^{\circ}\ Fe^{-(0.0112\ min^{-1})t}\\ \frac{20^{\circ}\ F }{105^{\circ}\ F} &= e^{-(0.0112\ min^{-1})t}\\ \ln \bigg[ \frac{20}{105} &= e^{-(0.0112\ min^{-1})t}\bigg]\\ \ln \frac{20}{105} &= -(0.0112\ min^{-1})t \\ t &= \frac{1}{-0.0112\ min^{-1}}\ln \frac{20}{105} \\ t &= \boxed{ 148\ min} \end{align*} {/eq}

It will take around {eq}\boxed{ 148\ min} {/eq} for the turkey to cool to {eq}95^{\circ}\ F {/eq}.