Newton's Law of Cooling tells us that the rate of change of the temperature of an object is proportional to the temperature difference between the object and its surroundings. This can be modeled by the differential equation 
 dT/dt =k(T(t)−12) 
, where T 
 is the temperature of the object after  t units of time have passed,  
A is the ambient temperature of the object's surroundings, and  
K is a constant of proportionality.  Suppose that a cup of coffee begins at  187
 degrees and, after sitting in room temperature of 70 
 degrees for  15 
 minutes, the coffee reaches 177 
 degrees. How long will it take before the coffee reaches 157 
 degrees?

To solve this problem, we can use **Newton's Law of Cooling** and the given differential equation:

\[
\frac{dT}{dt} = k(T(t) - 12)
\]

However, there seems to be a slight mismatch in the provided equation. Based on Newton's Law of Cooling, the equation should use the ambient temperature instead of the constant \(12\), and the differential equation should be written as:

\[
\frac{dT}{dt} = k(T(t) - A)
\]

where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( A \) is the ambient temperature (here, 70°F),
- \( k \) is the constant of proportionality.

So, the correct differential equation should be:

\[
\frac{dT}{dt} = k(T(t) - 70)
\]

### Step 1: Solve the Differential Equation

This is a separable differential equation. To solve it, we first rewrite it as:

\[
\frac{dT}{T(t) - 70} = k \, dt
\]

Now, integrate both sides:

\[
\int \frac{1}{T(t) - 70} \, dT = \int k \, dt
\]

This gives:

\[
\ln|T(t) - 70| = kt + C
\]

Exponentiating both sides:

\[
T(t) - 70 = e^{kt + C} = e^C \cdot e^{kt}
\]

Let \( C_1 = e^C \), so:

\[
T(t) = 70 + C_1 e^{kt}
\]

### Step 2: Apply Initial Conditions to Find Constants

We know that at \( t = 0 \), the initial temperature of the coffee is 187°F. Using this, we can find \( C_1 \):

\[
T(0) = 187 = 70 + C_1 e^{k \cdot 0}
\]

\[
C_1 = 187 - 70 = 117
\]

So, the temperature function becomes:

\[
T(t) = 70 + 117 e^{kt}
\]

### Step 3: Use the Data to Find \( k \)

We are given that after 15 minutes, the temperature is 177°F. Substituting \( T(15) = 177 \) into the equation:

\[
177 = 70 + 117 e^{15k}
\]

\[
107 = 117 e^{15k}
\]

\[
\frac{107}{117} = e^{15k}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{107}{117}\right) = 15k
\]

\[
k = \frac{\ln\left(\frac{107}{117}\right)}{15}
\]

Let's compute \( k \):

\[
k \approx \frac{\ln(0.91453)}{15} \approx \frac{-0.0896}{15} \approx -0.00597
\]

### Step 4: Determine the Time When Temperature Reaches 157°F

Now we want to find the time \( t \) when the temperature is 157°F. Substitute \( T(t) = 157 \) into the equation:

\[
157 = 70 + 117 e^{-0.00597t}
\]

\[
87 = 117 e^{-0.00597t}
\]

\[
\frac{87}{117} = e^{-0.00597t}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{87}{117}\right) = -0.00597t
\]

\[
t = \frac{\ln\left(\frac{87}{117}\right)}{-0.00597}
\]

Now let's compute \( t \):

\[
t \approx \frac{\ln(0.74359)}{-0.00597} \approx \frac{-0.2963}{-0.00597} \approx 49.63
\]

Thus, it will take approximately **49.63 minutes** for the coffee to reach 157°F.

### Summary
- The coffee will take about **49.63 minutes** to cool down to 157°F.

Would you like more details or further clarification?

### 5 Related Questions
1. How does Newton's Law of Cooling apply to real-life scenarios, such as cooling food or drinks?
2. How can the value of \( k \) be influenced by external factors like insulation or air flow?
3. What happens if the ambient temperature changes over time?
4. Can we model other heat transfer problems using Newton’s Law of Cooling?
5. How accurate is Newton's Law of Cooling for non-homogeneous materials?

### Tip
For better results in cooling predictions, consider external factors like airflow or container insulation, as they can affect the rate of heat loss.

Newton's Law of Cooling tells us that the rate of change of the temperature of an object is proportional to the temperature difference between the object and its surroundings. This can be modeled by the differential equation dT/dt = k(T(t) − 12), where T is the temperature of the object after t units of time have passed, A is the ambient temperature of the object's surroundings, and K is a constant of proportionality. Suppose that a cup of coffee begins at 187 degrees and, after sitting in room temperature of 70 degrees for 15 minutes, the coffee reaches 177 degrees. How long will it take before the coffee reaches 157 degrees?

Learn how to apply Newton's Law of Cooling to find how long it takes a cup of coffee to cool from 187°F to 157°F in a room with a constant temperature of 70°F. This problem explores differential equations and exponential decay, suitable for college-level calculus students.

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