The temperature of a cooling body follows an exponential decay. If the initial temperature is 70°C and it decreases to 55°C after 15 minutes, what will he the temperature after 42 mins?

To determine the temperature of a cooling body following exponential decay, we can use Newton's Law of Cooling, which states:

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

where:
- \( T(t) \) is the temperature at time \( t \),
- \( T_{\text{env}} \) is the ambient temperature,
- \( T_0 \) is the initial temperature,
- \( k \) is the cooling constant,
- \( t \) is the time.

Given:
- Initial temperature \( T_0 = 70^\circ \text{C} \),
- Temperature after 15 minutes \( T(15) = 55^\circ \text{C} \),
- Time \( t = 42 \) minutes.

First, we need to determine the cooling constant \( k \). Since we do not know the ambient temperature \( T_{\text{env}} \), we can simplify the problem assuming that the final temperature is significantly lower than the initial temperature, but typically this is given or can be inferred (e.g., room temperature). For simplicity and without loss of generality, let's assume the ambient temperature is 0°C (this will simplify calculations, but note that this is an assumption for theoretical calculation).

From the given information:
\[ T(15) = 55 = 70 e^{-15k} \]

To find \( k \):
1. Rearrange the equation to solve for \( e^{-15k} \):
   \[ 55 = 70 e^{-15k} \]
   \[ \frac{55}{70} = e^{-15k} \]
   \[ \frac{11}{14} = e^{-15k} \]

2. Take the natural logarithm of both sides:
   \[ \ln\left(\frac{11}{14}\right) = -15k \]

3. Solve for \( k \):
   \[ k = -\frac{1}{15} \ln\left(\frac{11}{14}\right) \]

   Calculate the natural logarithm and division:
   \[ k \approx -\frac{1}{15} \ln(0.7857) \]
   \[ k \approx -\frac{1}{15} \times (-0.2412) \]
   \[ k \approx 0.01608 \]

Next, use the value of \( k \) to determine the temperature after 42 minutes:
\[ T(42) = 70 e^{-0.01608 \times 42} \]

Calculate the exponent and the temperature:
\[ T(42) = 70 e^{-0.676} \]
\[ T(42) = 70 \times 0.508 \]
\[ T(42) \approx 35.56^\circ \text{C} \]

So, the temperature after 42 minutes will be approximately \( 35.56^\circ \text{C} \).

Would you like more details or have any questions?

Here are some related questions:
1. How does Newton's Law of Cooling apply to real-life scenarios?
2. What factors can affect the cooling rate of an object besides the ambient temperature?
3. How can we modify the formula if the ambient temperature is not 0°C?
4. What are the implications of different cooling constants \( k \) for different materials?
5. How would the temperature change if the object were in a different environment, such as a vacuum or a fluid?

**Tip:** Always consider the ambient temperature when solving cooling problems, as it can significantly impact the results.

Explore how to solve exponential decay in a cooling body using Newton's Law of Cooling. Starting from an initial temperature of 70°C, the temperature decreases to 55°C after 15 minutes. Find out what the temperature will be after 42 minutes with a detailed step-by-step solution. Suitable for high school students studying physics or mathematics.

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