​Newton's law of cooling says that the rate at which a body cools is proportional to the difference in temperature between the body and an environment into which it is introduced. This leads to an equation where the temperature​ f(t) of the body at time t after being introduced into an environment having constant temperature 

Upper T 0T0 

is 

f left parenthesis t right parenthesis equals Upper T 0 plus Upper C e Superscript negative kt Baseline commaf(t)=T0+Ce−kt, 

where C and k are constants. If 

Upper C equals 80 commaC=80, 

k equals 0.1 commak=0.1, 

and t is time in​ minutes, how long will it take a hot cup of coffee to cool to a temperature of 

2626degrees Upper C°C 

in a room at 

1818degrees Upper C°C​?

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.

Learn how to calculate the time it takes for a hot cup of coffee to cool to 26°C using Newton's Law of Cooling. This problem involves understanding exponential cooling processes and applying the formula f(t) = T_0 + C e^{-kt}.

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