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 and is placed on a table in a room where the temperature is . If is the temperature of the turkey after t minutes, then Newton's Law of Cooling implies that . This could be solved as a separable differential equation. Another method is to make the change of variable . If the temperature of the turkey is after half an hour, what is the temperature after 35 min?

Question

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  and is placed on a table in a room where the temperature is  . If  is the temperature of the turkey after t minutes, then Newton's Law of Cooling implies that  . This could be solved as a separable differential equation. Another method is to make the change of variable  . If the temperature of the turkey is  after half an hour, what is the temperature after 35 min?

Quizplus · Accepted Answer

Using the change of variable $u=T-T_a$, we have $\frac{dT}{dt} = \frac{du}{dt}$ and the differential equation becomes $\frac{du}{dt} = -k(u+T_a-11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11)$. This is a separable differential equation, and we can solve it by separating variables and integrating:
$$\int_{u_0}^{u(t)} \frac{du}{u+T_a-11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11} = -k \int_0^t dt$$
The left-hand side can be evaluated using the natural logarithm:
$$\ln \left|\frac{u(t)+T_a-11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11}{u_0+T_a-11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11}ight| = -kt$$
Solving for $u(t)$ and using the fact that $T(t)=u(t)+T_a$, we obtain:
$$T(t) = (T_0-T_a)e^{-kt}+T_a+11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11$$
Plugging in $T_0=11eaa3e5\_552f\_1b0a\_9f8f\_0392095402bc\_TB5971\_11$, $T_a=11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11$, and $t=35$ minutes, we find:
$$T(35) = (11eaa3e5\_552f\_1b0a\_9f8f\_0392095402bc\_TB5971\_11-11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11)e^{-k\cdot 35}+11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11+11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11$$
Since we are given that $T(30) = 11eaa3e5\_552f\_421f\_9f8f\_1fc1b245db67\_TB5971\_11$, we can use this to solve for $k$:
$$11eaa3e5\_552f\_421f\_9f8f\_1fc1b245db67\_TB5971\_11 = (11eaa3e5\_552f\_1b0a\_9f8f\_0392095402bc\_TB5971\_11-11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11)e^{-k\cdot 30}+11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11+11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11$$
Solving for $k$, we find $k \approx 0.0487$ min$^{-1}$. Plugging this back into the formula for $T(t)$, we obtain:
\begin{align*}
T(35) &= (11eaa3e5\_552f\_1b0a\_9f8f\_0392095402bc\_TB5971\_11-11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11)e^{-0.0487\cdot 35}\
&+11eaa3e5\_552f\_1b0b\_9f8f\_11cb69df6062\_TB5971\_11+11eaa3e5\_552f\_1b0b\_9f8f\_0392095402bc\_TB5971\_11\
&\approx 11eaa3e5\_552f\_4221\_9f8f\_416ad3af9d57\_TB5971\_11
\end{align*}
Therefore, the correct answer is $\boxed{	extbf{(B)}}$.

Newton's Law of Cooling States That the Rate of Cooling