This problem involves Newton's Law of Cooling, which describes the rate at which an object changes temperature when it is in contact with its surrounding environment. The law can be expressed as:

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

Where:

• $T(t)$ is the temperature of the object at time $t$,
• $T_s$ is the surrounding temperature,
• $k$ is the constant of proportionality,
• $t$ is the time.

Step-by-step solution:

We are given the following information:

• Initial temperature of water: $T(0) = 100^\circ \text{C}$,
• Temperature of the water after 10 minutes: $T(10) = 80^\circ \text{C}$,
• Surrounding temperature: $T_s = 30^\circ \text{C}$,
• We are asked to find the time when the water reaches $T = 40^\circ \text{C}$.

Step 1: Set up the general solution to the differential equation.

The solution to Newton's Law of Cooling is:

$T(t) = T_s + (T_0 - T_s)e^{-kt}$

Where:

• $T_0$ is the initial temperature (100°C in this case),
• $T(t)$ is the temperature of the object at time $t$,
• $k$ is the cooling constant,
• $T_s$ is the surrounding temperature (30°C).

So the equation becomes:

$T(t) = 30 + (100 - 30)e^{-kt}$ $T(t) = 30 + 70e^{-kt}$

Step 2: Use the given information to find $k$.

We are told that after 10 minutes, the water has cooled to $80^\circ \text{C}$. Plugging this into the equation:

$80 = 30 + 70e^{-10k}$

Simplifying:

$80 - 30 = 70e^{-10k}$ $50 = 70e^{-10k}$ $\frac{50}{70} = e^{-10k}$ $\frac{5}{7} = e^{-10k}$

Taking the natural logarithm of both sides:

$\ln\left(\frac{5}{7}\right) = -10k$ $k = -\frac{1}{10} \ln\left(\frac{5}{7}\right)$ $k \approx 0.02899$

Step 3: Find the time when the temperature is $40^\circ \text{C}$.

Now, we want to find the time when the temperature reaches $T = 40^\circ \text{C}$. Plugging this into the equation:

$40 = 30 + 70e^{-kt}$ $40 - 30 = 70e^{-kt}$ $10 = 70e^{-kt}$ $\frac{10}{70} = e^{-kt}$ $\frac{1}{7} = e^{-kt}$

Taking the natural logarithm again:

$\ln\left(\frac{1}{7}\right) = -kt$ $kt = -\ln\left(\frac{1}{7}\right)$ $t = \frac{\ln(7)}{k}$

Substitute the value of $k \approx 0.02899$:

$t = \frac{\ln(7)}{0.02899}$ $t \approx \frac{1.9459}{0.02899}$ $t \approx 67.1 \text{ minutes}$

Final Answer:

The temperature of the water will reach $40^\circ \text{C}$ after approximately 67.1 minutes.

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Would you like any further details or have any other questions on this problem?

Here are some related questions:

1. How does the value of the surrounding temperature $T_s$ affect the cooling rate?
2. How would you derive the formula for Newton's Law of Cooling from first principles?
3. What other physical processes can be modeled by similar exponential decay equations?
4. What is the significance of the cooling constant $k$, and how can it be experimentally determined?
5. How does the rate of cooling change if the surrounding temperature is closer to the temperature of the object?

Tip: The cooling constant $k$ can vary with factors such as the material properties of the object and the medium through which heat is transferred. Experimentally determining $k$ involves measuring the object's temperature over time and fitting the data to the cooling equation.