To solve this problem using Newton's Law of Cooling, we will follow the formula [tex]\( f(t) = T_0 + C e^{-k t} \)[/tex].
Given:
- Environmental temperature, [tex]\( T_0 = 0^{\circ} F \)[/tex]
- Initial temperature of the coffee, [tex]\( T_{\text{initial}} = 140^{\circ} F \)[/tex]
- Temperature of the coffee after [tex]\( 15 \)[/tex] minutes, [tex]\( T_{\text{after 15}} = 41^{\circ} F \)[/tex]
- Time at [tex]\( t = 15 \)[/tex] minutes
- We need to find the temperature of the coffee after [tex]\( 20 \)[/tex] minutes
Let's solve step-by-step:
1. Determine the constant [tex]\( C \)[/tex]:
Using the initial temperature:
[tex]\[ T_{\text{initial}} = T_0 + C \][/tex]
[tex]\[ 140 = 0 + C \][/tex]
[tex]\[ C = 140 \][/tex]
2. Find the decay constant [tex]\( k \)[/tex]:
Using the temperature after 15 minutes:
[tex]\[ T_{\text{after 15}} = T_0 + C e^{-k \cdot 15} \][/tex]
[tex]\[ 41 = 0 + 140 e^{-15k} \][/tex]
[tex]\[ 41 = 140 e^{-15k} \][/tex]
[tex]\[ \frac{41}{140} = e^{-15k} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left(\frac{41}{140}\right) = -15k \][/tex]
[tex]\[ k = -\frac{\ln\left(\frac{41}{140}\right)}{15} \][/tex]
After solving, we find:
[tex]\[ k \approx 0.08187 \][/tex]
3. Calculate the temperature after 20 minutes:
Using the formula:
[tex]\[ T_{\text{after 20}} = T_0 + C e^{-k \cdot 20} \][/tex]
[tex]\[ T_{\text{after 20}} = 0 + 140 e^{-0.08187 \cdot 20} \][/tex]
Simplifying:
[tex]\[ T_{\text{after 20}} \approx 27.227 \][/tex]
Rounding to the nearest integer, we get:
[tex]\[ T_{\text{after 20}} \approx 27^{\circ} F \][/tex]
Hence, the coffee’s temperature after 20 minutes is [tex]\( 27^{\circ} F \)[/tex].
