# Newton's Law of Cooling, find room temperature

## 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-Hs)=(H0-Hs)ekt
dH/dt = -k(H-Hs)

Hs 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.

Last edited:

ehild
Homework Helper
You have the temperatures at 10 minutes and at 20 minutes. The exponential terms are e10k and e20k which is the square of e10k . Plug in the temperature data and use that the second equation is just the square of the first one.

ehild

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-Hs), the equation above is derived from it.

It didn't lead to an unfactorable polynomial when I tried it. . . I ended up substituting M=e10k, which made it easier to solve (or less intimidating, anyways).

ehild
Homework Helper
It works, just try.

Let be z=e10k. Then e20k=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

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.