Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Suppose that a roast turkey is taken from an oven when its temperature has reached $160 ^ { \circ } \mathrm { F }$ and is placed on a table in a room where the temperature is $60 ^ { \circ } \mathrm { F }$. If $u ( t )$ is the temperature of the turkey after $t$ minutes, then Newton's Law of Cooling implies that $\frac { d u } { d t } = k ( u - 60 ) \text {. }$ This could be solved as a separable differential equation. Another method is to make the change of variable $y = u - 60$. If the temperature of the turkey is $150 ^ { \circ } \mathrm { F }$ after half an hour, what is the temperature after $35 \mathrm {~min}$ ?
A) $t = 143 ^ { \circ } \mathrm { F }$
B) $t = 148 ^ { \circ } \mathrm { F }$
C) $t = 298 ^ { \circ } \mathrm { F }$
D) none of these
E) $t = 95 ^ { \circ } \mathrm { F }$

Question

Quizplus · Accepted Answer

We are given that $\frac { d u } { d t } = k ( u - 60 ) $ and that 150 degrees Fahrenheit is the temperature of turkey after half an hour, i.e., $u(30) = 150$.

Let $y = u - 60$. Then $\frac { d y } { d t } = \frac { d u } { d t } = k ( u - 60 ) = k y$.

This is a separable differential equation, and we can use separation of variables to obtain:

$$\frac { d y } { y } = k dt $$

Integrating both sides, we get:

$$\ln |y| = kt + C$$

where $C$ is the constant of integration.

Using the initial condition $u(30) = 150$ or $y(30) = 90$, we get:

$$\ln |90| = 30k + C$$.

Simplifying, we obtain:

$$C = \ln 90 - 30k$$

Substituting this value of $C$ in the equation above, we get:

$$\ln |y| = kt + \ln 90 - 30k$$

Further simplification gives:

$$|y| = e ^ { kt } e ^ { \ln 90-30k } = Ke^{kt}$$

where $K$ is a positive constant given by $K=|y(0)|=|u(0) - 60|$.

We are now given that we need to find the temperature after 35 minutes, i.e., we need to find $u(35)$, or equivalently, $y(35)+60$.

Substituting $t = 35$ and using the value for $K$ obtained above, we get:

$$e ^ { 35k } K = e ^ { 35k } |y(0)| = |y(35)|$$

Substituting this value of $|y(35)|$ in the equation $y = u - 60$, we get:

$$u(35) = y(35) + 60 = e ^ { 35k } |y(0)| + 60 = Ke^{35k} + 60.$$

We can now use the initial condition given in the problem, namely, that the temperature is $160$ degrees Fahrenheit initially, to obtain the value of $K$ in terms of $k$.

We have:

$$K = |u(0) - 60| = |160 - 60| = 100$$

Substituting this value of $K$ in the expression for $u(35)$, we get:

$$u(35) = Ke^{35k}+60 = 100e^{35k}+60$$

We are given that the temperature of turkey is $150 ^ { \circ } \mathrm { F }$ after half an hour, i.e., $u(30) = 150$. Substituting this in the above equation, we get:

$$150 = 100e^{30k}+60$$

Solving for $k$, we get:

$$k = \frac { 1 } { 30 } \ln \left( \frac { 9 } { 5 } \right)$$

Substituting this value of $k$ in the expression for $u(35)$, we get:

$$u ( 35 ) = 100 \cdot \mathrm { e } ^ { \frac { 7 } { 6 } \mathrm { ln } \left( \frac { 9 } { 5 } \right) } + 60 = 148 ^ { \circ } \mathrm { F } $$

Therefore, the temperature of the turkey after 35 minutes is $148 ^ { \circ } \mathrm { F } $, which is choice (B).

Newton's Law of Cooling States That the Rate of Cooling 160∘F160 ^ { \circ } \mathrm { F }160∘F

Newton's Law of Cooling States That the Rate of Cooling $160 ^ { \circ } \mathrm { F }$