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The image contains a math question, part 2(b), that states:

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**2. (b)** A body originally at 80°C cools down to 60°C in 20 minutes, the temperature of the air being 40°C. Find the temperature of the body after 40 minutes from the original. (07 Marks)

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This is a classic problem related to **Newton's Law of Cooling**, which states that the rate of change of the temperature of a body is proportional to the difference between its current temperature and the ambient temperature.

Mathematically, Newton's Law of Cooling can be written as:

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

Where:
- \( T \) is the temperature of the body at time \( t \),
- \( T_{\text{ambient}} \) is the ambient temperature (in this case, 40°C),
- \( k \) is a constant that depends on the cooling process.

We are given the following:
- Initial temperature of the body: \( T(0) = 80°C \),
- After 20 minutes: \( T(20) = 60°C \),
- Ambient temperature: \( T_{\text{ambient}} = 40°C \).

We need to find the temperature of the body after 40 minutes: \( T(40) \).

### Steps to Solve:

1. First, we use the formula:
   \[
   T(t) = T_{\text{ambient}} + (T(0) - T_{\text{ambient}}) e^{-kt}
   \]
   Substituting \( T_{\text{ambient}} = 40°C \), and \( T(0) = 80°C \), we get:
   \[
   T(t) = 40 + 40 e^{-kt}
   \]

2. We are given that \( T(20) = 60°C \). Substituting this into the equation:
   \[
   60 = 40 + 40 e^{-20k}
   \]
   Solving for \( k \):
   \[
   20 = 40 e^{-20k} \implies \frac{1}{2} = e^{-20k}
   \implies -20k = \ln\left(\frac{1}{2}\right)
   \implies k = \frac{\ln(2)}{20}
   \]

3. Now, substitute this value of \( k \) back into the equation to find \( T(40) \):
   \[
   T(40) = 40 + 40 e^{-40k}
   \]
   Using \( k = \frac{\ln(2)}{20} \), this simplifies to:
   \[
   T(40) = 40 + 40 e^{-40 \cdot \frac{\ln(2)}{20}} = 40 + 40 e^{-2\ln(2)} = 40 + 40 \cdot \left(\frac{1}{2^2}\right)
   \]
   \[
   T(40) = 40 + 40 \cdot \frac{1}{4} = 40 + 10 = 50°C
   \]

Thus, the temperature of the body after 40 minutes will be **50°C**.

Would you like further explanation on any of the steps or details?

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Here are five related questions to deepen your understanding:
1. How does the value of \( k \) affect the cooling process in Newton's Law of Cooling?
2. What happens to the temperature if the ambient temperature is lower than the body’s initial temperature?
3. Can Newton's Law of Cooling be applied to heating problems? If so, how?
4. How would the solution change if the ambient temperature was not constant but varied with time?
5. What is the physical meaning of the exponential function in the cooling formula?

**Tip**: Always remember that Newton's Law of Cooling is based on the assumption that the ambient temperature remains constant over time.

A body originally at 80°C cools down to 60°C in 20 minutes, the temperature of the air being 40°C. Find the temperature of the body after 40 minutes from the original.

This problem involves solving a Newton's Law of Cooling equation, where a body cools from 80°C to 60°C in 20 minutes, and the air temperature is constant at 40°C. Learn the steps to find the temperature after 40 minutes using an exponential decay formula.

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