To solve this problem, let's work with the given differential equation:

$T(t) = 106e^{\left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t} + 66$

where:

• $T(t)$ is the temperature of the coffee at time $t$,
• $T(0) = 172^\circ F$,
• the coffee cools to $165^\circ F$ after $14$ minutes in a room temperature of $66^\circ F$,
• and we want to find the time $t$ when the coffee temperature $T(t)$ reaches $154^\circ F$.

Step-by-Step Solution

1. Set up the equation for $T(t) = 154$: $154 = 106e^{\left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t} + 66$

2. Isolate the exponential term: $154 - 66 = 106e^{\left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t}$ $88 = 106e^{\left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t}$

3. Divide by 106: $\frac{88}{106} = e^{\left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t}$

4. Take the natural logarithm of both sides to solve for $t$: $\ln \left(\frac{88}{106}\right) = \left(\frac{1}{14} \ln \left(\frac{99}{106}\right)\right)t$

5. Isolate $t$: $t = \frac{\ln \left(\frac{88}{106}\right)}{\frac{1}{14} \ln \left(\frac{99}{106}\right)}$

6. Calculate $t$:

   • First, find $\ln \left(\frac{88}{106}\right)$.
   • Then, find $\ln \left(\frac{99}{106}\right)$.
   • Divide these results to get $t$.

After performing the calculations, you will get:

$t \approx 33.62 \text{ minutes}$

So, it will take approximately 33.62 minutes for the coffee to cool to $154^\circ F$.

Would you like more details or have any questions?

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Related Questions:

1. How does the room temperature affect the cooling rate of the coffee?
2. How would the solution change if the initial temperature of the coffee was different?
3. How does Newton's Law of Cooling apply to this problem?
4. What would be the temperature of the coffee after 45 minutes?
5. If the room temperature increased, how would that affect the time to reach 154 degrees?

Tip:

When dealing with exponential decay or growth problems, carefully isolate the variable using logarithmic properties.