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The problem in the image relates to Newton's Law of Cooling. Here's a transcription of the problem:

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**b)** The temperature of the body drops from 100°C to 75°C in 10 minutes when the surrounding air is at 20°C. What will be its temperature after half an hour? When will the temperature be 25°C?  
*(07 Marks each)*

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This can be solved using **Newton's Law of Cooling**, which is given by the formula:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_s \) is the surrounding (ambient) temperature.
- \( T_0 \) is the initial temperature of the object.
- \( k \) is the cooling constant (which we will calculate).
- \( t \) is the time in minutes.

### Step 1: Calculate the Cooling Constant \( k \)
From the given dataThe problem in the image relates to Newton's Law of Cooling. Here's a transcription of the problem:

---

**b)** The temperature of the body drops from 100°C to 75°C in 10 minutes when the surrounding air is at 20°C. What will be its temperature after half an hour? When will the temperature be 25°C?  
*(07 Marks each)*

---

This can be solved using **Newton's Law of Cooling**, which is given by the formula:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_s \) is the surrounding (ambient) temperature.
- \( T_0 \) is the initial temperature of the object.
- \( k \) is the cooling constant (which we will calculate).
- \( t \) is the time in minutes.

### Step 1: Calculate the Cooling Constant \( k \)
From the given 
- Initial temperature \( T_0 = 100^\circ C \)
- Ambient temperature \( T_s = 20^\circ C \)
- After 10 minutes, the temperature \( T(10) = 75^\circ C \).

Substituting these values into the Newton’s Law of Cooling equation:

\[
75 = 20 + (100 - 20) e^{-10k}
\]

Simplify the equation to solve for \( k \).

### Step 2: Temperature after 30 minutes
After finding \( k \), you can substitute \( t = 30 \) minutes into the cooling equation to find the temperature at that time.

### Step 3: Time when temperature reaches 25°C
For this, use the equation \( T(t) = 25^\circ C \) and solve for \( t \).

Would you like me to proceed with the full calculation to determine \( k \), the temperature after 30 minutes, and the time when the temperature reaches 25°C?

Let me know if you want details or have any questions!

Here are 5 related questions to think about:
1. How does the value of \( k \) change if the surrounding temperature is higher?
2. What would happen if the body were placed in an environment colder than 20°C?
3. Can this equation be used for heating instead of cooling?
4. How does Newton’s Law of Cooling apply to real-world problems like cooling of a hot drink?
5. How accurate is Newton's Law of Cooling for objects with large mass?

**Tip:** When solving exponential decay problems, it's helpful to remember that natural logarithms (\(\ln\)) are often used to isolate the time variable.

The temperature of the body drops from 100°C to 75°C in 10 minutes when the surrounding air is at 20°C. What will be its temperature after half an hour? When will the temperature be 25°C?

Solve the problem of temperature drop using Newton's Law of Cooling. Calculate the temperature of a body after half an hour when the surrounding air is 20°C. Learn about exponential decay and cooling constants in this advanced problem, suitable for Grades 11-12.

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