Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the surrounding medium. Thus, if an object is taken from an oven at 297F and left to cool in a room at 72F, its temperature T
 after t
 hours will satisfy the differential equation dT/dt= k(T-72)
 
If the temperature fell to 200
F in 0.8 hours, what will it be after 5 hours? After 5 hours, the temperature will be 
 
F.

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.

Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference in temperature between the object and the surrounding medium. Thus, if an object is taken from an oven at 297F and left to cool in a room at 72F, its temperature T after t hours will satisfy the differential equation dT/dt = k(T - 72). If the temperature fell to 200F in 0.8 hours, what will it be after 5 hours?

Solve a cooling problem using Newton's Law of Cooling. Learn how to calculate the temperature of an object after 5 hours using a differential equation. Initial temperature is 297°F, room temperature is 72°F, and the temperature falls to 200°F in 0.8 hours. Suitable for advanced high school students (Grades 11-12).

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