To solve this problem, we'll use Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature. The law is mathematically represented as:

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

Where:

• $T$ is the temperature of the object at time $t$.
• $T_{\text{ambient}}$ is the ambient (room) temperature.
• $k$ is a positive constant that depends on the properties of the object.
• $t$ is the time in minutes.

Given:

• Initial temperature, $T(0) = 8^\circ C$
• Room temperature, $T_{\text{ambient}} = 30^\circ C$
• Temperature after 3 minutes, $T(3) = 13^\circ C$

Let's solve step by step:

Step 1: Find the general solution for $T(t)$

Newton's Law of Cooling can be integrated to give the formula:

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

Where:

• $T_0$ is the initial temperature of the thermometer (8°C in this case).

So, the equation becomes:

$T(t) = 30 + (8 - 30) e^{-kt}$

Simplifying:

$T(t) = 30 - 22 e^{-kt}$

Step 2: Find $k$ using the given data

We know that $T(3) = 13^\circ C$:

$13 = 30 - 22 e^{-3k}$

Rearranging this equation:

$22 e^{-3k} = 30 - 13 = 17$

Thus:

$e^{-3k} = \frac{17}{22}$

Taking the natural logarithm of both sides:

$-3k = \ln \left(\frac{17}{22}\right)$

$k = -\frac{1}{3} \ln \left(\frac{17}{22}\right)$

Calculating this value:

$k \approx 0.088$

Step 3: Find the temperature after 5 and 10 minutes

Now we can plug the values of $k$ and $t$ into the equation $T(t) = 30 - 22 e^{-0.088t}$.

Temperature after 5 minutes:

$T(5) = 30 - 22 e^{-0.088 \times 5}$

Calculating:

$T(5) \approx 30 - 22 e^{-0.44} \approx 30 - 22 \times 0.644 \approx 30 - 14.17 \approx 15.83^\circ C$

Temperature after 10 minutes:

$T(10) = 30 - 22 e^{-0.088 \times 10}$

Calculating:

$T(10) \approx 30 - 22 e^{-0.88} \approx 30 - 22 \times 0.415 \approx 30 - 9.13 \approx 20.87^\circ C$

Step 4: Graph the relationship between temperature and time

The graph of $T(t) = 30 - 22 e^{-0.088t}$ will show an exponential curve starting at $8^\circ C$ and approaching $30^\circ C$ as $t$ increases.

To verify these calculations using a graphing tool, plot the function and use TRACE to check the temperatures at $t = 5$ and $t = 10$.

Would you like a detailed graph plotted, or any further explanation?

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5 Related Questions:

1. How would the result change if the ambient temperature were different?
2. What happens to the rate of temperature change as the thermometer's temperature approaches room temperature?
3. Can you derive the formula for Newton's Law of Cooling from first principles?
4. How does the constant $k$ affect the speed of cooling or heating?
5. What are some practical applications of Newton's Law of Cooling?

Tip:

Newton's Law of Cooling can also be applied in reverse for heating problems where the object is initially cooler than its surroundings.