# A roast turkey is taken from the oven when its temperature has reached 185 degrees Fand is placed...

## Question:

A roast turkey is taken from the oven when its temperature has reached 185 degrees Fand is placed on a table in a room where the temperature is 75 degrees F. If the temperature of the turkey is 150 degrees F after half an hour, what is the temperature after 70 minutes?

## Newton's Cooling Law

Newton's cooling law equation is a type of an exponential function that is used to describe the temperature of an object placed in an environment with constant temperature. Thermodynamics dictates that the temperature of the object must change over time such that it is in thermal equilibrium with the surrounding or it should have the same temperature with the surrounding.

The temperature of the turkey at any time t is given by Newton's cooling law equation,

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

where {eq}T_r = 75^{\circ} \text{ F} {/eq} is the temperature of the room, {eq}T_i = 185^{\circ} \text{ F} {/eq} is the initial temperature of the turkey and k is the cooling constant to be calculated.

To solve the problem, we need to solve for k and write an expression for the temperature of the turkey over time. The constant k can be solved using the fact that it takes {eq}30 \text{ min } {/eq} for the turkey to cool to {eq}150^{\circ}\ \text{F} {/eq}. We have, {eq}T(30 \text{ min}) = 150^{\circ} \text{ F} {/eq} and plug it into the general equation.

{eq}\displaystyle \begin{align*} T(t) &= T_r + (T_i - T_r)e^{-kt},\ T(30 \text{ min}) = 150^{\circ} \text{ F}\\ 150^{\circ} \text{ F} &= 75^{\circ} \text{ F} + (185^{\circ} \text{ F}- 75^{\circ} \text{ F})e^{-k(30 \text{ min}) } \\ 150^{\circ} \text{ F} - 75^{\circ} \text{ F} &= (185^{\circ} \text{ F}- 75^{\circ} \text{ F})e^{-k(30 \text{ min}) } \\ 75^{\circ} \text{ F} &= 110^{\circ} \text{ F}e^{-k(30 \text{ min}) } \\ e^{-k(30 \text{ min}) } &= \frac{75^{\circ} \text{ F}}{110^{\circ} \text{ F}} \\ \text{Taking the natural logarithm of both sides: }\\ \ln \bigg[ e^{-k(30 \text{ min}) } &= \frac{75^{\circ} \text{ F}}{110^{\circ} \text{ F}} \bigg] \\ -k(30 \text{ min}) &= \ln \frac{75}{110} \\ k &= \frac{1}{-30 \text{ min}}\ln \frac{75}{110} \\ k &\approx \boxed{ 0.012766\text{ min}^{-1}} \end{align*} {/eq}

Now we can express the Newton's cooling law equation for this problem as {eq}\displaystyle \boxed{ T(t) = 75^{\circ} \text{ F} + 110^{\circ}\text{ F} e^{-(0.012766\text{ min}^{-1})t}}. {/eq}

(a.) If we want to solve the temperature of the turkey after 70 minutes, we plug {eq}t = 70 \text{ min } {/eq} into {eq}T(t) {/eq}.

{eq}\displaystyle \begin{align*} T(t) &= 75^{\circ} \text{ F} + 110^{\circ} \text{ F}e^{-(0.012766\text{ min}^{-1})t},\ t = 70 \text{ min} \\ T(70 \text{ min} ) &= 75^{\circ} \text{ F} + 110^{\circ} \text{ F}e^{-(0.012766\text{ min}^{-1})(70 \text{ min})}\\ &= 75^{\circ} \text{ F} + 45.0^{\circ} \text{ F} \\ T(70 \text{ min} ) &= \boxed{120.0^{\circ} \text{ F} } \end{align*} {/eq}

The temperature of the turkey after 70 minutes is {eq}\boxed{120.0^{\circ} \text{ F} } {/eq}.