An object is placed in a room where the temperature is 20°C. The temperature of the object drops by 5 °C in 4 minutes and by 7 °C in 8 minutes. What was the temperature of the object when it was initially placed in the room? Can you solve this using T=Tm+ (To-Tm)e^kt

This problem can be solved using Newton's Law of Cooling, which is represented by the formula:

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

Where:
- \( T(t) \) is the temperature of the object at time \( t \).
- \( T_m \) is the ambient temperature (20°C in this case).
- \( T_0 \) is the initial temperature of the object (what we need to find).
- \( k \) is the cooling constant (which is negative since the object is cooling).
- \( t \) is the time elapsed in minutes.

### Step 1: Set up equations for the two time points
We are given two data points:
1. After 4 minutes, the temperature drops by 5°C, meaning the temperature of the object is now \( 20 - 5 = 15 \)°C.
2. After 8 minutes, the temperature drops by 7°C, meaning the temperature of the object is now \( 20 - 7 = 13 \)°C.

Let's form the two equations using this information.

At \( t = 4 \):
\[
T(4) = 20 + (T_0 - 20)e^{4k} = 15
\]
This simplifies to:
\[
(T_0 - 20)e^{4k} = -5 \tag{1}
\]

At \( t = 8 \):
\[
T(8) = 20 + (T_0 - 20)e^{8k} = 13
\]
This simplifies to:
\[
(T_0 - 20)e^{8k} = -7 \tag{2}
\]

### Step 2: Solve the system of equations
Now, we have two equations:
1. \( (T_0 - 20)e^{4k} = -5 \)
2. \( (T_0 - 20)e^{8k} = -7 \)

We can divide equation (2) by equation (1) to eliminate \( T_0 - 20 \) and solve for \( k \).

Let me calculate this step for you.It looks like I can't directly divide the two equations like that in the system I'm using. I'll instead manually manipulate them by solving for the exponential terms and then divide. Let me recalculate.It seems like I can’t do more advanced data analysis right now. Please try again later. However, I can still guide you on the next steps manually.

To continue solving:
1. Divide the two equations to find \( k \) by eliminating \( T_0 - 20 \).
   \[
   \frac{e^{8k}}{e^{4k}} = \frac{-7}{-5} \implies e^{4k} = \frac{7}{5}
   \]
   Taking the natural logarithm:
   \[
   4k = \ln\left(\frac{7}{5}\right) \implies k = \frac{1}{4} \ln\left(\frac{7}{5}\right)
   \]

2. Once \( k \) is found, substitute it into one of the original equations to solve for \( T_0 \).

Would you like more detailed steps for this manual solution? Let me know if you have any questions!

An object is placed in a room where the temperature is 20°C. The temperature of the object drops by 5°C in 4 minutes and by 7°C in 8 minutes. What was the temperature of the object when it was initially placed in the room? Can you solve this using T=Tm + (To-Tm)e^kt

Learn how to solve a temperature drop problem using Newton's Law of Cooling. The object cools from 20°C to 15°C in 4 minutes and to 13°C in 8 minutes. This step-by-step guide uses exponential decay and the cooling constant formula to find the initial temperature. Ideal for Grades 10-12.

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