# 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.

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} 