A bowl of soup is initially at 90 degrees Celsius and is placed in a room with a constant temperature of 20 degrees Celsius. After 8 minutes, the soup cools down to 72 degrees Celsius. Using Newton's Law of Cooling, determine the temperature of the soup after 14 minutes.

Newton's Law of Cooling describes the rate of cooling of an object in relation to the temperature of its surroundings. The law is expressed as:

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

where:
- \( T \) is the temperature of the object (the soup) at time \( t \),
- \( T_s \) is the surrounding temperature (20°C in this case),
- \( k \) is a constant that depends on the properties of the object and its environment,
- \( t \) is the time,
- \( \frac{dT}{dt} \) is the rate of change of temperature.

The solution to this differential equation can be written as:

\[
T(t) = T_s + (T_0 - T_s) e^{-kt}
\]

where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_0 \) is the initial temperature of the object.

### Step 1: Plugging in the known values
- Initial temperature \( T_0 = 90^\circ C \),
- Surrounding temperature \( T_s = 20^\circ C \),
- Temperature after 8 minutes \( T(8) = 72^\circ C \).

Using these values, we can first solve for \( k \).

\[
72 = 20 + (90 - 20) e^{-8k}
\]
\[
72 = 20 + 70 e^{-8k}
\]
\[
52 = 70 e^{-8k}
\]
\[
e^{-8k} = \frac{52}{70} = 0.74286
\]
\[
-8k = \ln(0.74286)
\]
\[
k = -\frac{\ln(0.74286)}{8} \approx 0.03617
\]

### Step 2: Solving for the temperature after 14 minutes
Now that we have \( k \approx 0.03617 \), we can determine the temperature at \( t = 14 \) minutes.

\[
T(14) = 20 + (90 - 20) e^{-14 \cdot 0.03617}
\]
\[
T(14) = 20 + 70 e^{-0.50638}
\]
\[
T(14) = 20 + 70 \cdot 0.60258
\]
\[
T(14) = 20 + 42.18
\]
\[
T(14) \approx 62.18^\circ C
\]

### Conclusion
The temperature of the soup after 14 minutes is approximately **62.18°C**.

Would you like more details or have any questions?

Here are 5 related questions:
1. How does Newton's Law of Cooling change if the room temperature varies over time?
2. What other factors, besides temperature difference, could influence the cooling rate of an object?
3. How can you estimate the constant \( k \) from experimental data in different scenarios?
4. How long would it take for the soup to cool to 40°C under the same conditions?
5. Can Newton's Law of Cooling be used to model heating processes as well?

**Tip:** The value of \( k \) is key to determining how fast an object cools, and it can vary based on the material and environment.

Determine the temperature of a bowl of soup using Newton's Law of Cooling. The soup starts at 90°C in a room at 20°C, cooling to 72°C after 8 minutes. Learn how to calculate the temperature after 14 minutes using key mathematical concepts like exponential decay and differential equations. Suitable for Grades 10-12.

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