an object is heated 90 degrees and allowed to cool in  a room with constant ambient temperature of 20 degrees. If after 10 minutes the temperature of the object is 60 degrees, what will its temperature be after 20 minutes?

This problem is based on **Newton's Law of Cooling**, which states that the rate of cooling of an object is proportional to the difference between the temperature of the object and the ambient temperature. Mathematically, it is expressed as:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_{\text{ambient}} \) is the ambient temperature,
- \( T_0 \) is the initial temperature of the object,
- \( k \) is a constant that depends on the material and cooling conditions,
- \( t \) is the time elapsed.

### Given information:
- Initial temperature \( T_0 = 90^\circ \text{C} \),
- Ambient temperature \( T_{\text{ambient}} = 20^\circ \text{C} \),
- After 10 minutes, the temperature is \( T(10) = 60^\circ \text{C} \).

First, we will use this information to find the cooling constant \( k \).

### Step 1: Find the constant \( k \)

From the formula:

\[
T(10) = 20 + (90 - 20) e^{-10k}
\]

Substitute \( T(10) = 60 \):

\[
60 = 20 + 70 e^{-10k}
\]

Simplify the equation:

\[
40 = 70 e^{-10k}
\]

\[
\frac{40}{70} = e^{-10k}
\]

\[
\frac{4}{7} = e^{-10k}
\]

Take the natural logarithm of both sides:

\[
\ln\left(\frac{4}{7}\right) = -10k
\]

Now solve for \( k \):

\[
k = -\frac{1}{10} \ln\left(\frac{4}{7}\right)
\]

### Step 2: Find the temperature after 20 minutes

Now that we have \( k \), we can use it to find the temperature after 20 minutes.

\[
T(20) = 20 + 70 e^{-20k}
\]

Let me calculate both \( k \) and \( T(20) \).The temperature of the object after 20 minutes will be approximately \( 42.86^\circ \text{C} \).

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

Here are 5 related questions:
1. How would the temperature change if the ambient temperature were different?
2. Can Newton's Law of Cooling be used for heating processes as well?
3. What would the temperature be after 30 minutes?
4. How can the cooling constant \( k \) change depending on different materials?
5. How is this model limited in real-world applications?

**Tip:** Newton's Law of Cooling assumes that the temperature difference is the primary factor influencing heat transfer, but real-world factors (like air flow) can alter the cooling rate.

An object is heated 90 degrees and allowed to cool in a room with constant ambient temperature of 20 degrees. If after 10 minutes the temperature of the object is 60 degrees, what will its temperature be after 20 minutes?

Learn how to calculate the temperature of an object after 20 minutes using Newton's Law of Cooling. The object was initially heated to 90°C, and after 10 minutes, its temperature dropped to 60°C in a room with a constant ambient temperature of 20°C. This step-by-step solution includes calculus and exponential decay concepts. Suitable for Grades 11-12.

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