Differential equation question

A bath is filled with water which is allowed to cool. The temperature of the water is x degrees celsius after cooling for t minutes and the temperature of the room is constant at 20 degrees celsius.
Given that the rate at which the temperature of the water decreases is proportional to the difference in temperature between the water and the room

1. Write down a differential equation connecting x and t.

Given also that the temperature of the water is initially 37 degrees and that it is 36 degrees after cooling for four minutes.
2. Find the temperature of the water after ten minutes.

For part one I get dx/dt=k(x-20) -is this correct?

And for part two what should I get as the final equation and what is the correct answer? What should I get for the value of k?

I will rep for a good response. Thanks.

Reply 1

Jonny W

8

V1000

For part one I get dx/dt=k(x-20) -is this correct?

Yes. I would say dx/dt = -k(x - 20) to emphasise that the constant in front of (x - 20) is negative. But that is not essential, and I will use your differential equation in my solution.

1/(x - 20) dx/dt = k . . . . . separating the variables
ln(x - 20) = c + kt . . . . . integrating wrt t

Now we find the constants c and k.

ln(17) = c . . . . . since x = 37 when t = 0
ln(16) = c - 4k . . . . . since x = 36 when t = 4

k
= (1/4)(ln(16) - ln(17))
= (1/4)ln(16/17)

Taking e^ of each side in

ln(x - 20) = ln(17) + (1/4)ln(16/17) t

gives

x - 20 = 17e^((1/4)ln(16/17) t)
x = 20 + 17e^((1/4)ln(16/17) t)

When t = 10, x = 20 + 17e^((1/4)ln(16/17)*10) = 34.609.

[A good check on the answer is that 34.609 is slightly more than 37 - 2.5. Tell me the reason and I'll rep you!]

Reply 2

dvs

10

Your equation is correct. (Note that, because of the way you set it up, we should expect a negative value for k. This is because the temperature is decreasing.)

So, let's solve the thing:
dx/dt = k(x-20)
dx/(x-20) = k dt
ln(x-20) = kt + C
x = 20 + Ae^(kt), where A=e^C.

When t=0, x=37, and when t=4, x=36. So:
37 = 20 + A => A = 17
36 = 20 + 17e^(4k) => k = 0.25 ln(16/17)

Therefore:
x = 20 + 17e^(0.25 ln(16/17) t)

Setting t=10 will give us the temperature after 10mins.

Edit:
Looks like I was beaten to it. :p:

Reply 3

Malik

12

Jonny is it because temp dropped from 1degree in 4minutes, therefore you can assume it would roughly drop (1/4)t in t minutes. So if it started at 37 after 10minutes you would expect temp to be around 37-2.5?

Reply 4

dvs

10

Malik

Jonny is it because temp dropped from 1degree in 4minutes, therefore you can assume it would roughly drop (1/4)t in t minutes. So if it started at 37 after 10minutes you would expect temp to be around 37-2.5?

I would think that was his reasoning. It's looking at the exponential decay from a linear perspective. (Try plotting the graphs of x=-0.25t+37 and the actual equation on the same axes to see this more clearly.)

Reply 5

Malik

12

dvs

I would think that was his reasoning. It's looking at the exponential decay from a linear perspective. (Try plotting the graphs of x=-0.25t+37 and the actual equation on the same axes to see this more clearly.)

lol ill take your word for it, actually just because I am bored Ill check it out :wink:

edit/ did it just for first ten mins, you can see difference increasing as it goes on.

tempgraph.GIF5.1KB

Differential equation question

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