# The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where is the temperature of the coffee, A is

The temperature of a cup of coffee varies according to Newton’s Law of Cooling: -“dT/dt=k(T-A), where is the temperature of the coffee, A is the room temperature, and k is a positive
constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25*C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes,
74
67
60
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B) 67°C.

Step-by-step explanation:

Newton’s Law of Cooling is given by:

$$\displaystyle \frac{dT}{dt}=k(T-A)$$

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 (TA). Hence:

$$\displaystyle \frac{dT}{T-A}=k\, dt$$

Take the integral of both sides:

$$\displaystyle \int \frac{dT}{T-A}=\int k\, dt$$

Integrate:

$$\displaystyle \ln\left|T-A\right| = kt+C$$

Raise both sides to e:

$$|T-A|=e^{kt+C}=Ce^{kt}$$

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:

$$\displaystyle T=Ce^{kt}+A$$

We are given that A = 25. Hence:

$$\displaystyle T=Ce^{kt}+25$$

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:

$$100=Ce^{k(0)}+25\Rightarrow C=75$$

Hence:

$$T=75e^{kt}+25$$

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:

$$90=75e^{k(1)}+25$$

Solve for k:

$$\displaystyle e^k=\frac{13}{15}\Rightarrow k=\ln\left(\frac{13}{15}\right)$$

Therefore:

$$\displaystyle T=75e^{\ln({}^{13}\! /\!{}_{15})t}+25$$

Then after four minutes, the temperature of the coffee will be:

\displaystyle \begin{aligned} \displaystyle T&=75e^{\ln({}^{13}\! /\!{}_{15})(4)}+25\\\\&\approx 67^\circ\text{C}\end{aligned}