When some food is taken from a refrigerator at 𝑡=0t=0 minutes, its temperature is 1∘𝐶1∘C. After 𝑡=3t=3 minutes in a 25∘𝐶25∘C room its temperature has increased to 9∘𝐶9∘C. What is the temperature of the food at 𝑡=10t=10 minutes?

This problem involves Newton's Law of Cooling, which can also describe heating in this case. The formula for Newton's Law of Cooling is:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_{\text{room}} \) is the room temperature (25°C in this case).
- \( T_{\text{initial}} \) is the initial temperature of the object (1°C in this case).
- \( k \) is a constant that depends on the characteristics of the object.
- \( t \) is the time in minutes.

We are given:
- \( T(0) = 1^\circ C \),
- \( T(3) = 9^\circ C \),
- \( T_{\text{room}} = 25^\circ C \).

### Step 1: Use the data at \( t = 3 \) to find \( k \)

We begin by using the formula to express \( T(3) \):

\[
9 = 25 + (1 - 25) e^{-3k}
\]

Simplifying:

\[
9 = 25 - 24 e^{-3k}
\]

\[
9 - 25 = -24 e^{-3k}
\]

\[
-16 = -24 e^{-3k}
\]

\[
e^{-3k} = \frac{2}{3}
\]

Taking the natural logarithm of both sides:

\[
-3k = \ln\left( \frac{2}{3} \right)
\]

\[
k = -\frac{1}{3} \ln\left( \frac{2}{3} \right)
\]

### Step 2: Use \( k \) to find the temperature at \( t = 10 \)

Now we can substitute \( k \) back into the equation to find the temperature at \( t = 10 \).

The equation becomes:

\[
T(10) = 25 + (1 - 25) e^{-10k}
\]

Let me compute the values to find the final result.The temperature of the food at \( t = 10 \) minutes is approximately \( 18.79^\circ C \).

Would you like more details, or do you have any questions?

Here are some related questions to explore further:
1. How does Newton's Law of Cooling apply to different materials?
2. What happens if the initial and room temperatures are reversed in this problem?
3. How would the cooling process differ if the room temperature changes over time?
4. Can Newton’s Law of Cooling apply to objects with irregular shapes?
5. How would the result change if we considered heat transfer by radiation?

**Tip:** Newton's Law of Cooling assumes that the temperature change is proportional to the difference between the object's temperature and the surrounding environment. It works best for small temperature differences.

When some food is taken from a refrigerator at t=0 minutes, its temperature is 1°C. After t=3 minutes in a 25°C room, its temperature has increased to 9°C. What is the temperature of the food at t=10 minutes?

Solve a Newton's Law of Cooling problem involving food taken from a refrigerator at 1°C. After 3 minutes in a 25°C room, the temperature increases to 9°C. Learn how to find the temperature after 10 minutes using exponential decay. Suitable for Grades 10-12.

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