# A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed...

## Question:

A roast turkey is taken from an oven when its temperature has reached 185 degrees F and is placed on a table in a room where

the temperature is 75 degrees F.

(a) If the temperature of the turkey is 150 degrees F after half an hour, what is the temperature after 55 minutes?

(b) When will the turkey have cooled to 100 degrees F?

## Newton's Law of Cooling

Newton's law of cooling states that the rate of heat loss of a system is directly proportional to the difference in the temperature between the system and the surroundings.

## Answer and Explanation:

We can solve the problem using the Newton's law of cooling, which is given by

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

where {eq}T(t) {/eq} is the temperature at time {eq}t {/eq}, {eq}T_s {/eq} is the temperature of the room, {eq}T_0 {/eq} is the initial temperature of the roast turkey, and

{eq}k {/eq} is a constant. After substituting the given values, equation becomes

{eq}T(t)=75+(185-75)e^{-kt}\\ T(t)=75+110e^{-kt} {/eq}

A. We must solve first for the value of k using the condition {eq}T(30\,\rm min)=150^\circF {/eq}

{eq}150=75+110e^{-k{30}}\\ \dfrac{75}{110}=e^{-30k}\\ \text{Note that } \ln e^x = x\\ \ln \mid 0.68 \mid = \ln e^{-30k}\\ k=\dfrac{-0.383}{-30}\\ k\approx 0.0128 \\ {/eq}

Therefore, after 55 minutes, temperature of the roast turkey would be

{eq}T(55)\approx 75+110e^{-0.0128(55)}\\ T(55)\approx 75+54.5\\ \color{blue}{T(55)\approx 129.5^\circ F} {/eq}

B. The time until the turkey have cooled to {eq}100^\circ F {/eq}

would be

{eq}100=75+110e^{-0.0128t}\\ \dfrac{25}{110}=e^{-0.0128t}\\ \ln \mid 0.227 \mid \approx -0.0128t\\ \dfrac{-1.48}{-0.0128}\approx t\\ \color {blue}{t\approx 115.75\,\rm min} {/eq}