# Newton's Law of Cooling of porridge

1. Oct 17, 2007

### denverhockeyfan

1. The problem statement, all variables and given/known data
A smaller bowl of porridge served at 200 degres F cools to 160 degres in 1 min. What tempature (too cold) will this porridge be when the bowl of exercise 27 has reached 120 degres F (just right)?

2. Relevant equations
y(prime)(t)=k((y(t)-T sub a )
y(t)=Ae^(kt)+T sub a

3. The attempt at a solution
I really don't get this.
Thanks for your help, and if you can explain that would be great.

2. Oct 17, 2007

### PiratePhysicist

Well, this is just a simple differential equation:

$$y' = \frac{dT}{dt}=k (T-T_a)$$
$$\frac{dT}{(T-T_a)}=k dt$$
$$\int{\frac{dy}{(T-T_a)}},\int{k dt}$$
$$ln(T-T_a) = k t + C$$
$$T=Ae^{kt}+T_a$$
So that's how the two equations are related.
Now:
$$T(0)=Ae^{k 0}+T_a=T_0$$ Where $$T_0$$ is the temperature the soup starts at (initial T) and $$T_a$$ is the temp of the air around the soup(ambient T). Use 200 degrees as the starting temp.
$$T(t)=(T_0-T_a)e^{kt}+T_a$$
So now you need the constant k. Plug in 1 min for t, 160 for T(t) and then the values for the initial T and the ambient T and solve for k. After all that you can plug the time from execercise 27 and find the temp of the soup.