Question

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Suppose tt is time, T is the temperature of the object, and Ts is the surrounding temperature. The following differential equation describes Newton's Law dTdt=k(T−Ts),where k is a constant. Suppose that we consider a 90∘C cup of coffee in a 24∘C room. Suppose it is known that the coffee cools at a rate of 1∘C/min. when it is 70∘C. Answer the following questions.
1. Find the constant k in the differential equation.
Answer (in per minute): k=

2. What is the limiting value of the temperature?
Answer (in Celsius): T=

3. Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes.
Answer (in Celsius): T(10)≈

Answer 1

Answer:

1) $k=-\frac{1}{46}\approx -0.02$ 

2) $\textit{Limiting value} = 24$

3) $T(10)\approx \frac{494611944}{6436343}\approx 76.8$

Explanation:

First of all note that  

$T_s=24$ is the surroundings temperature, the temperature of the room where the cup of coffee is. Then, the differential equation is:  

$\frac{dT}{dt}=k(T-24)$

Also, note that all units are in degrees celsius and minutes. Then, we don't have to convert units. Let's not write units explicitly from now on.  

Explanation  

1) We have that  

$\textit{rate of cooling}=\frac{dT}{dt}=1,\quad T=70$

at some point - the exact time at which this is true doesn't really play any role because the equation doesn't have t on the right hand side. Then, from the equation we get  

$1=-(70-24)=46k\Rightarrow k=-\frac{1}{46}\approx -0.02$

The minus comes from considering the temperature must decrease. With this value we can write the equation more explicitely:  

$\frac{dT}{dt}=-\frac{1}{46}(T-24)$

2) The coffee is cooling off as time goes by, and it won't get any cooler than 24 degrees celsius because that's the temperature of the room. Then, in the long run, the temperature of the coffee is 24 degrees celsius.  

3) Remember that Euler's method consists of using an initial exact measurement to predict what will happen in the future, approximately. There is a formula to make those predictions an it depends on the time step they gave us. Let's compute things first and then I tell you the equations we used.  

In this case we know that we start with a 90 degrees celsius cup of coffee, or, in terms of math,  

$T(0)=90$

Then, we can predict:  

$T(2)\approx 90+2\left[-\frac{1}{46}(90-24)\right]=\frac{2004}{23}\approx 87.1$

Let's use fractions so we don't lose accuracy from now. With this number we can make an approximation of the temperature after 2 more seconds:  

$T(4)\approx \frac{2004}{23}+2\left[-\frac{1}{46}\left(\frac{2004}{23}-24\right)\right]=\frac{44640}{529}\approx 84.4$

and then  

$T(6)\approx \frac{44640}{529}+2\left[-\frac{1}{46}\left(\frac{44640}{529}-24\right)\right]=\frac{994776}{12167}\approx 81.8$

and then  

$T(8)\approx \frac{994776}{12167}+2\left[-\frac{1}{46}\left(\frac{994776}{12167}-24\right)\right]=\frac{22177080}{279841}\approx 79.2$  

and finally, the number we wanted to find:

$T(10)\approx \frac{22177080}{279841}+2\left[-\frac{1}{46}\left(\frac{22177080}{279841}-24\right)\right]=\frac{494611944}{6436343}\approx 76.8$  

I hope you noticed the pattern to compute the next prediction:  

$\textit{next prediction} = \textit{previous one (or exact value if it's the first step)}\\+ h\ast(\textit{right hand side of the differential equation at the previous one})$