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This problem is based on **Newton's Law of Cooling**, which states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the ambient temperature.

Let:
- \( T(t) \) be the temperature of the object at time \( t \) (in minutes),
- \( T_r \) be the constant room temperature (\( 10^\circ C \)),
- \( T_0 \) be the initial temperature of the object (\( 100^\circ C \)),
- \( k \) be the cooling constant.

### Part (a): Finding \( T(t) \)

According to Newton's Law of Cooling, we have the differential equation:
\[
\frac{dT}{dt} = -k(T(t) - T_r)
\]
Substituting the given values, \( T_r = 10 \):
\[
\frac{dT}{dt} = -k(T(t) - 10)
\]
This is a separable differential equation. To solve it:

1. **Separate variables**:
\[
\frac{dT}{T - 10} = -k \, dt
\]

2. **Integrate both sides**:
\[
\int \frac{1}{T - 10} \, dT = -k \int dt
\]
This gives:
\[
\ln|T - 10| = -kt + C
\]
Exponentiating both sides:
\[
T - 10 = e^{-kt + C} = A e^{-kt}
\]
where \( A = e^C \).

3. **General solution**:
\[
T(t) = 10 + A e^{-kt}
\]

4. **Initial condition**: When \( t = 0 \), \( T(0) = 100 \):
\[
100 = 10 + A \implies A = 90
\]
Thus, the solution becomes:
\[
T(t) = 10 + 90 e^{-kt}
\]

5. **Using the second condition**: After 10 minutes, the temperature is \( 60^\circ C \):
\[
60 = 10 + 90 e^{-10k}
\]
Simplifying:
\[
50 = 90 e^{-10k} \implies e^{-10k} = \frac{5}{9}
\]
Taking the natural logarithm:
\[
-10k = \ln \left( \frac{5}{9} \right) \implies k = -\frac{1}{10} \ln \left( \frac{5}{9} \right)
\]
This gives the value of \( k \). Now, we compute:
\[
k \approx 0.04577
\]

Thus, the temperature function is:
\[
T(t) = 10 + 90 e^{-0.04577t}
\]

### Part (b): Time to reach \( 30^\circ C \)

We need to find \( t \) when \( T(t) = 30 \). Using the equation:
\[
30 = 10 + 90 e^{-0.04577t}
\]
Simplifying:
\[
20 = 90 e^{-0.04577t} \implies e^{-0.04577t} = \frac{2}{9}
\]
Taking the natural logarithm:
\[
-0.04577t = \ln \left( \frac{2}{9} \right) \implies t = -\frac{\ln \left( \frac{2}{9} \right)}{0.04577}
\]
This gives:
\[
t \approx 24.03 \text{ minutes}
\]

### Part (c): Behavior as \( t \to \infty \)

As \( t \to \infty \), \( e^{-0.04577t} \to 0 \). Therefore:
\[
T(t) \to 10
\]
This means that as time goes on, the temperature of the object will approach the room temperature, which is \( 10^\circ C \).

### Summary:
- **(a)** \( T(t) = 10 + 90 e^{-0.04577t} \)
- **(b)** The object will reach \( 30^\circ C \) in approximately 24.03 minutes.
- **(c)** As \( t \to \infty \), the temperature will approach \( 10^\circ C \).

Would you like more details or have any questions?

#### Related Questions:
1. How does changing the room temperature affect the cooling rate?
2. Can you derive Newton's Law of Cooling from basic principles?
3. What are the assumptions behind Newton's Law of Cooling?
4. How can you modify the equation if the room temperature fluctuates over time?
5. How would the cooling rate change if the object was insulated?

#### Tip:
Always check units carefully when solving problems involving time, temperature, and rates of change to ensure consistency throughout calculations.

A metal object at a temperature of 100°C is placed in a room which has a constant temperature of 10°C. The object is cooling at a rate proportional to the difference between its temperature and the room temperature. After 10 minutes the temperature of the object is 60°C. (a) Find the function T that gives the temperature of the object at time t. (b) Find the time it will take for the object to reach a temperature of 30°C (c) What happens as t → ∞? Use mathematics to support your answer.

Learn how to apply Newton's Law of Cooling to find the temperature of a metal object placed in a room. The problem explores solving differential equations, finding exponential decay functions, and determining behavior as time approaches infinity. Suitable for students in Grades 11-12 and college-level studies.

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