When some food is taken from a refrigerator at 𝑡=0
 minutes, its temperature is 1∘𝐶
. After 𝑡=2
 minutes in a 20∘𝐶
 room its temperature has increased to 8∘𝐶
. What is the temperature of the food at 𝑡=8
 minutes?

We can solve this problem using **Newton's Law of Cooling**, which states that the rate of change of the temperature of an object is proportional to the difference between the temperature of the object and the surrounding environment. Mathematically, it is given by:

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

Where:
- \(T(t)\) is the temperature of the object at time \(t\),
- \(T_s\) is the surrounding temperature (20°C in this case),
- \(T_0\) is the initial temperature of the object (1°C at \(t=0\)),
- \(k\) is the cooling (or heating) constant,
- \(t\) is the time in minutes.

### Step 1: Use the given information at \( t = 2 \)
At \( t = 2 \) minutes, the temperature \( T(2) = 8^\circ C \). Using the equation, we can plug in values to find \(k\).

\[
8 = 20 + (1 - 20)e^{-2k}
\]
\[
8 = 20 - 19e^{-2k}
\]
\[
-12 = -19e^{-2k}
\]
\[
e^{-2k} = \frac{12}{19}
\]
Now, take the natural logarithm on both sides to solve for \(k\):

\[
-2k = \ln\left(\frac{12}{19}\right)
\]
\[
k = -\frac{1}{2} \ln\left(\frac{12}{19}\right)
\]

### Step 2: Find the temperature at \( t = 8 \)
Once we find \( k \), we can use it to find the temperature at \( t = 8 \). The general equation becomes:

\[
T(8) = 20 + (1 - 20)e^{-8k}
\]

Let's calculate this step by step.The temperature of the food at \( t = 8 \) minutes is approximately \( 16.98^\circ C \).

Would you like more details or have any other questions?

Here are five related questions:
1. How does Newton's Law of Cooling apply to cooling scenarios instead of heating?
2. How can the cooling constant \(k\) vary with different environmental conditions?
3. What would happen if the surrounding temperature was lower than the initial temperature of the object?
4. How does Newton's Law of Cooling model temperature changes over very long periods of time?
5. How can this law be applied to practical situations, like cooling food or beverages quickly?

**Tip:** Newton's Law of Cooling assumes a constant surrounding temperature, but in reality, the surroundings may change, leading to more complex models.

When some food is taken from a refrigerator at t=0 minutes, its temperature is 1°C. After t=2 minutes in a 20°C room its temperature has increased to 8°C. What is the temperature of the food at t=8 minutes?

Use Newton's Law of Cooling to determine the temperature of food taken from a refrigerator at 1°C after 8 minutes in a 20°C room. Learn the application of differential equations and exponential functions in this step-by-step solution. Ideal for Grades 10-12.

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