Given the problem, we have the equation from Newton's Law of Cooling:

$f(t) = T_0 + C e^{-kt}$

where:

• $f(t)$ is the temperature of the coffee at time $t$.
• $T_0 = 18^\circ C$ is the constant room temperature.
• $C = 80$ is the initial temperature difference between the coffee and the room.
• $k = 0.1$ is the cooling constant.
• $t$ is the time in minutes.

We need to find the time $t$ when the coffee cools to $f(t) = 26^\circ C$.

Step 1: Set up the equation with the given values.

$26 = 18 + 80 \cdot e^{-0.1t}$

Step 2: Solve for $t$.

1. Subtract 18 from both sides: $8 = 80 \cdot e^{-0.1t}$

2. Divide both sides by 80: $\frac{8}{80} = e^{-0.1t}$

   $0.1 = e^{-0.1t}$

3. Take the natural logarithm of both sides: $\ln(0.1) = -0.1t$

4. Solve for $t$: $t = \frac{\ln(0.1)}{-0.1}$

Let's calculate this value of $t$.The coffee will take approximately 23.03 minutes to cool down to $26^\circ C$ in a room at $18^\circ C$.

Would you like any further details or have any questions?

Related Questions:

1. How long would it take for the coffee to cool to $20^\circ C$?
2. What is the temperature of the coffee after 10 minutes?
3. How does changing the value of $k$ affect the cooling process?
4. What would be the initial temperature of the coffee if it started at $95^\circ C$?
5. How would the cooling time change if the room temperature was $22^\circ C$?

Tip: Newton's Law of Cooling can also be applied to heating processes, where a body warms up towards the ambient temperature instead of cooling down.