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.