Newton’s law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton’s law of cooling. If the coffee has a temperature of 200˝F when freshly poured, and 60 seconds later has cooled to 190˝F in a room at 70˝F, determine when the coffee reaches a temperature of 150˝F.

Newton's Law of Cooling is typically expressed mathematically as:

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

where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_s \) is the surrounding temperature,
- \( k \) is a positive constant related to the cooling rate,
- \( t \) is the time elapsed.

### Step 1: Solve the differential equation

The differential equation can be solved by separating variables and integrating. Rearranging the equation:

\[
\frac{dT}{T - T_s} = -k \, dt
\]

Integrating both sides:

\[
\int \frac{1}{T - T_s} \, dT = -k \int dt
\]

\[
\ln |T - T_s| = -kt + C
\]

Where \( C \) is the integration constant. Exponentiating both sides:

\[
|T - T_s| = e^{-kt + C} = e^C \cdot e^{-kt}
\]

Let \( e^C = C_1 \):

\[
T(t) - T_s = C_1 e^{-kt}
\]

So:

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

### Step 2: Determine the constant \( C_1 \) and \( k \)

When \( t = 0 \), \( T(0) = 200^\circ F \):

\[
200 = 70 + C_1 e^{-k \cdot 0}
\]

\[
C_1 = 200 - 70 = 130
\]

So the temperature function is:

\[
T(t) = 70 + 130 e^{-kt}
\]

Using the information that \( T(60) = 190^\circ F \):

\[
190 = 70 + 130 e^{-60k}
\]

\[
120 = 130 e^{-60k}
\]

\[
\frac{120}{130} = e^{-60k}
\]

\[
\frac{12}{13} = e^{-60k}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{12}{13}\right) = -60k
\]

\[
k = -\frac{1}{60} \ln\left(\frac{12}{13}\right)
\]

Now calculate \( k \):

\[
k \approx -\frac{1}{60} \ln\left(0.9231\right)
\]

\[
k \approx \frac{1}{60} \times 0.0800 = 0.00133
\]

### Step 3: Determine the time when \( T(t) = 150^\circ F \)

Now use the temperature equation to find the time \( t \) when \( T(t) = 150^\circ F \):

\[
150 = 70 + 130 e^{-0.00133t}
\]

\[
80 = 130 e^{-0.00133t}
\]

\[
\frac{8}{13} = e^{-0.00133t}
\]

Taking the natural logarithm of both sides:

\[
\ln\left(\frac{8}{13}\right) = -0.00133t
\]

\[
t = -\frac{\ln\left(\frac{8}{13}\right)}{0.00133}
\]

Finally, calculating \( t \):

\[
t \approx \frac{\ln\left(1.625\right)}{0.00133} \approx \frac{0.4855}{0.00133} \approx 365 \text{ seconds}
\]

### Conclusion:

The coffee will reach a temperature of \( 150^\circ F \) after approximately **365 seconds** (or about **6 minutes and 5 seconds**).

Would you like more details or have any questions?

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**Related Questions:**

1. How does the value of \( k \) affect the rate of cooling?
2. Can Newton's Law of Cooling be applied to heating processes as well?
3. What are some real-world applications of Newton's Law of Cooling?
4. How would the solution change if the room temperature varied over time?
5. What would happen if the initial temperature was closer to the surrounding temperature?

**Tip:** To better understand the cooling process, consider visualizing the cooling curve using graphing software.

Learn how to apply Newton’s Law of Cooling to determine temperature change over time. In this example, calculate when a cup of coffee cools from 200˝F to 150˝F in a room at 70˝F. Detailed step-by-step solution included.

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