Answer:
B) 67°C.
Step-by-step explanation:
Newton's Law of Cooling is given by:
[tex]\displaystyle \frac{dT}{dt}=k(T-A)[/tex]
Where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.
We are given that the coffee cools from 100°C to 90°C in one minute at a room temperature A of 25°C.
And we want to find the temperature of the coffee after four minutes.
First, solve the differential equation. Multiply both sides by dt and divide both sides by (T - A). Hence:
[tex]\displaystyle \frac{dT}{T-A}=k\, dt[/tex]
Take the integral of both sides:
[tex]\displaystyle \int \frac{dT}{T-A}=\int k\, dt[/tex]
Integrate:
[tex]\displaystyle \ln\left|T-A\right| = kt+C[/tex]
Raise both sides to e:
[tex]|T-A|=e^{kt+C}=Ce^{kt}[/tex]
The temperature of the coffee T will always be greater than or equal to the room temperature A. Thus, we can remove the absolute value:
[tex]\displaystyle T=Ce^{kt}+A[/tex]
We are given that A = 25. Hence:
[tex]\displaystyle T=Ce^{kt}+25[/tex]
Since the coffee cools from 100°C to 90°C, the initial temperature of the coffee was 100°C. Thus, when t = 0,T = 100:
[tex]100=Ce^{k(0)}+25\Rightarrow C=75[/tex]
Hence:
[tex]T=75e^{kt}+25[/tex]
We are given that the coffee cools from 100°C to 90°C after one minute at a room temperature of 25°C.
So, T = 90 given that t = 1. Substitute:
[tex]90=75e^{k(1)}+25[/tex]
Solve for k:
[tex]\displaystyle e^k=\frac{13}{15}\Rightarrow k=\ln\left(\frac{13}{15}\right)[/tex]
Therefore:
[tex]\displaystyle T=75e^{\ln({}^{13}\! /\!{}_{15})t}+25[/tex]
Then after four minutes, the temperature of the coffee will be:
[tex]\displaystyle \begin{aligned} \displaystyle T&=75e^{\ln({}^{13}\! /\!{}_{15})(4)}+25\\\\&\approx 67^\circ\text{C}\end{aligned}[/tex]
Hence, our answer is B.
