A bowl of soup is initially at 88 degrees Celsius and is placed in a room with a constant temperature of 19 degrees Celsius. After 7 minutes, the soup cools down to 73 degrees Celsius. Using Newton's Law of Cooling, determine the temperature of the soup after 8 minutes.

Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between the object's temperature and the ambient temperature. Mathematically, it is expressed as:

\[
\frac{dT}{dt} = -k(T - T_{\text{ambient}})
\]

Where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_{\text{ambient}} \) is the constant ambient temperature,
- \( k \) is a positive constant,
- \( t \) is the time in minutes.

The solution to this differential equation can be written as:

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

Where:
- \( T_0 \) is the initial temperature of the object,
- \( T(t) \) is the temperature at time \( t \),
- \( k \) is a cooling constant.

### Given:
- \( T_0 = 88^\circ \text{C} \) (initial temperature),
- \( T_{\text{ambient}} = 19^\circ \text{C} \),
- \( T(7) = 73^\circ \text{C} \) after 7 minutes.

We first need to determine the value of the cooling constant \( k \) using the temperature at 7 minutes.

### Step 1: Use \( T(7) = 73^\circ \) to find \( k \)

Substitute the values into the equation:
\[
73 = 19 + (88 - 19)e^{-7k}
\]
\[
73 = 19 + 69e^{-7k}
\]
\[
54 = 69e^{-7k}
\]
\[
e^{-7k} = \frac{54}{69} = 0.7826
\]
\[
-7k = \ln(0.7826)
\]
\[
k = -\frac{\ln(0.7826)}{7}
\]

Let's compute \( k \).The cooling constant \( k \) is approximately \( 0.0350 \, \text{min}^{-1} \).

### Step 2: Determine the temperature after 8 minutes

Now, using the value of \( k \), we can find the temperature of the soup after 8 minutes by applying the formula:

\[
T(8) = 19 + (88 - 19)e^{-0.0350 \times 8}
\]

Let's calculate the temperature after 8 minutes.The temperature of the soup after 8 minutes is approximately \( 71.14^\circ \text{C} \).

Would you like more details on how this was calculated or any clarification on Newton's Law of Cooling?

Here are five related questions to expand on this topic:
1. How does the value of the cooling constant \( k \) affect the rate of cooling?
2. What would the soup's temperature be after 15 minutes?
3. How long will it take for the soup to cool to 50°C?
4. Can Newton's Law of Cooling be applied to heating processes?
5. How does the ambient temperature influence the cooling rate?

**Tip**: The cooling rate depends heavily on the temperature difference between the object and its surroundings. As the object approaches the ambient temperature, it cools more slowly.

Solve a real-world cooling problem using Newton's Law of Cooling. Learn how to calculate the temperature of a soup after 8 minutes when it cools from 88°C to 73°C in 7 minutes. Detailed solution provided with calculus and exponential decay principles. Suitable for Grades 10-12.

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