Newton's Law of Cooling is described by the differential equation:

$\frac{dT}{dt} = k(T - T_s)$

Where:

• $T(t)$ is the temperature of the object at time $t$,
• $T_s$ is the surrounding temperature (in this case, 72°F),
• $k$ is the constant of proportionality,
• $T(0) = 297°F$ (initial temperature).

We are given:

• $T_s = 72°F$,
• $T(0) = 297°F$,
• $T(0.8) = 200°F$.

Step 1: Solve the differential equation

The equation can be solved by separation of variables:

$\frac{dT}{T - T_s} = k \, dt$

Integrating both sides:

$\ln|T - T_s| = kt + C$

Exponentiating both sides:

$T - T_s = e^{kt + C} = C_1 e^{kt}$

Thus:

$T(t) = T_s + C_1 e^{kt}$

Substitute $T(0) = 297$ and $T_s = 72$:

$297 = 72 + C_1 e^{0}$

$C_1 = 297 - 72 = 225$

Thus, the equation becomes:

$T(t) = 72 + 225 e^{kt}$

Step 2: Determine $k$

We use the fact that $T(0.8) = 200$ to find $k$.

$200 = 72 + 225 e^{0.8k}$

$128 = 225 e^{0.8k}$

$e^{0.8k} = \frac{128}{225} = 0.5689$

Taking the natural logarithm:

$0.8k = \ln(0.5689) \approx -0.564$

$k = \frac{-0.564}{0.8} \approx -0.705$

Step 3: Find the temperature after 5 hours

Now that we know $k \approx -0.705$, we can find the temperature after 5 hours:

$T(5) = 72 + 225 e^{-0.705 \times 5}$

Let's calculate this.

The temperature of the object after 5 hours will be approximately $78.63^\circ F$.

Would you like more details or have any questions?

Here are 5 related questions:

1. How does the cooling constant $k$ affect the rate of cooling?
2. What assumptions are made when applying Newton's Law of Cooling?
3. Can Newton's Law of Cooling be applied to heating as well?
4. How would the temperature change if the room temperature $T_s$ were different?
5. How can this formula be modified for different units, such as Celsius?

Tip: The rate of cooling is faster when the temperature difference between the object and the surroundings is larger.