Newton's Law of Cooling, find room temperature

[hulgy]

Homework Statement

Suppose that the temperature of a pan of warm water obeys Newton's law of cooling. The water (47 degrees Celsius) was put in a room and 10 minutes later the water's temperature was 40 degrees Celsius. After another 10 minutes, the temperature of the water was 34 degrees Celsius. Find the room's temperature.

Homework Equations

(H-H_s)=(H₀-H_s)e^kt
dH/dt = -k(H-H_s)

H_s is the room's temperature
H₀ is the initial temperature of the water
H is the temperature of the water at time t

The Attempt at a Solution

I'm rather at a loss at how to do this problem. I've tried setting up a system of equations and tried finding the room temperature, but that doesn't work out. So if algebra won't do it, then I'm assuming something calculus-y should. I'm just at a lost as what that is.

 

[ehild]

You have the temperatures at 10 minutes and at 20 minutes. The exponential terms are e^10k and e^20k which is the square of e^10k . Plug in the temperature data and use that the second equation is just the square of the first one.

ehild

 

[hulgy]

It doesn't work, it only leads to a polynomial that can't be factored.

If it helps at all Newton's law of cooling is dH/dt = -k(H-H_s), the equation above is derived from it.

 

[Strants]

It didn't lead to an unfactorable polynomial when I tried it. . . I ended up substituting M=e^10k, which made it easier to solve (or less intimidating, anyways).

 

[ehild]

It works, just try.

Let be z=e^10k. Then e^20k=z^2.

You have two equations:

40-Hs=(47-Hs)z
34-Hs=(47-Hs)z^2

What about squaring the first equation and dividing by the second one?


ehild

 

[hulgy]

OK..

40-Hs=(47-Hs)z
34-Hs=(47-Hs)z^2

I square the first equation
(40-Hs)^2=(47-Hs)^2*z^2

and divide by the second, so I get:
(40-Hs)^2/(34-Hs) = (47-Hs)

...FFFFFUUUUUUUUUUUUUUUUUUUUU

I was doing this all along, but I was writing some of my signs wrong... so yeah. Anyway thanks for the help, answer is -2.