Roast turkey is taken from an oven when its temperature has reached 185 degree F and is placed on...

Question:

Roast turkey is taken from an oven when its temperature has reached 185 {eq}^\circ {/eq}F and is placed on a table in a room where the temperature is 75 {eq}^\circ {/eq}F.

(a) If the temperature of the turkey is 150 {eq}^\circ {/eq}F after half an hour, what is the temperature after 40 minutes?

(b) When will the turkey have cooled to 105 {eq}^\circ {/eq}F?

Newton's Cooling Law

Newton's Cooling Law is an exponential function that describes the temperature of an object over time when the object is placed in an environment with a constant temperature. The temperature of the object will because of the temperature gradient between the object and the environment.

Newton's Cooling Law is given by the equation

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

where T(t) is the temperature at any time t, {eq}T_s {/eq} is the temperature of the surrounding, {eq}T_0 {/eq} is the initial temperature of the object and k is the decay constant.

To solve this problem we to write the explicit function for T(t) of the turkey by solving for k.

Solving for k, we have the following given:

{eq}\displaystyle \begin{align*} t &= 30\ min \\ T_s &= 75^{\circ}\ F \\ T_0 &= 185^{\circ}\ F \\ T(30) &= 150^{\circ}\ F \end{align*} {/eq}

{eq}\displaystyle \begin{align*} T(t) &= T_s + (T_0 - T_s) e^{-kt} \\ T(30) &= 75^{\circ}\ F + (185^{\circ}\ F - 75^{\circ}\ F) e^{-k(30\ min)} \\ 150^{\circ}\ F &= 75^{\circ}\ F + (185^{\circ}\ F - 75^{\circ}\ F) e^{-k(30\ min)} \\ 150^{\circ}\ F - 75^{\circ}\ F &= 110^{\circ}\ Fe^{-k(30\ min)} \\ \frac{75^{\circ}\ F }{110^{\circ}\ F} &= e^{-k(30\ min)} \\ \frac{15 }{22} &= e^{-k(30\ min)} \end{align*} {/eq}

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

{eq}\displaystyle \begin{align*} \ln \bigg( \frac{15 }{22} &= e^{-k(30\ min)} \bigg) \\ \ln \bigg( \frac{15 }{22} \bigg) &= -k(30\ min) \\ k&= -\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1} \\ k &\approx 0.012766 \end{align*} {/eq}

Now we can write the explicit form of Newton's cooling law as {eq}\displaystyle \boxed{ T(t) = 75^{\circ}\ F + (110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t}} {/eq}.

a.) To solve for the temperature after 40 minutes we set t = 40 mins and use the function we derived above.

{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + 110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \\ &= 75^{\circ}\ F + 110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)(40\ min)} \\ T(t) &= \boxed{ 141.01 ^{\circ}\ F} \end{align*} {/eq}

The temperature of the turkey after 40 minutes is {eq}\boxed{ 141.01 ^{\circ}\ F} {/eq}.

b.) To solve for the time it takes for the turkey to reach a temperature of {eq}105^{\circ}\ F {/eq} we set the {eq}T(t) = 105^{\circ}\ F {/eq} and solve for t.

{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ}\ F + 110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \\ 105^{\circ}\ F &= 75^{\circ}\ F + 110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \\ 105^{\circ}\ F - 75^{\circ}\ F &= 110^{\circ}\ F e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \\ \frac{30}{ 110} &= e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \\ \end{align*} {/eq}

Taking the natural logarithm to solve for t.

{eq}\displaystyle \begin{align*} \ln \bigg( \frac{30}{ 110} &= e^{\bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t} \bigg) \\ \ln \bigg( \frac{30}{ 110} \bigg) &= \bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)t\\ t&= \frac{\ln \bigg( \frac{30}{ 110} \bigg) }{ \bigg(\frac{1}{30}\ln \bigg( \frac{15 }{22} \bigg)\ min^{-1}\bigg)}\\ t &= \boxed{ 101.8\ mins} \end{align*} {/eq}

This means that it will take around {eq}\boxed{ 101.8\ mins } {/eq} for the turkey to cool to {eq}105^{\circ}\ F {/eq}.