Question

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
42

Answer 1

B) 67°C.

Explanation:

Newton's Law of Cooling is given by:

`\displaystyle (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 (T - A). Hence:

`\displaystyle (dT)/(T-A)=k\, dt`

Take the integral of both sides:

`\displaystyle \int (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=(13)/(15)\Rightarrow k=\ln\left((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}`

Hence, our answer is B.