Newton's Law of Cooling of porridge

[denverhockeyfan]

Homework Statement

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)?

Homework Equations

y(prime)(t)=k((y(t)-T sub a )
y(t)=Ae^(kt)+T sub a

The Attempt at a Solution

I really don't get this.
Thanks for your help, and if you can explain that would be great.

 

[PiratePhysicist]

Well, this is just a simple differential equation:

[tex] y' = \frac{dT}{dt}=k (T-T_a)[/tex]
[tex]\frac{dT}{(T-T_a)}=k dt[/tex]
[tex]\int{\frac{dy}{(T-T_a)}},\int{k dt}[/tex]
[tex]ln(T-T_a) = k t + C[/tex]
[tex]T=Ae^{kt}+T_a[/tex]
So that's how the two equations are related.
Now:
[tex]T(0)=Ae^{k 0}+T_a=T_0[/tex] Where [tex]T_0[/tex] is the temperature the soup starts at (initial T) and [tex]T_a[/tex] is the temp of the air around the soup(ambient T). Use 200 degrees as the starting temp.
[tex]T(t)=(T_0-T_a)e^{kt}+T_a[/tex]
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.

 

[ogb p]

I am familiar with Newton's Law of Cooling, which states that the rate of change of temperature of an object is proportional to the difference between its current temperature and the ambient temperature. In this case, the ambient temperature would be the temperature of the room in which the porridge is cooling.

Based on the information given, we can use the equation y(t) = Ae^(kt) + T sub a to solve for the temperature of the porridge at any given time (t). Here, y(t) represents the temperature of the porridge, A is a constant, k is the cooling rate constant, and T sub a is the ambient temperature.

Since we know that the porridge cooled from 200 degrees F to 160 degrees F in 1 minute, we can plug in these values to solve for the constant k. This would give us the equation:

160 = Ae^(k*1) + T sub a

Next, we can use the given information that the porridge should reach 120 degrees F at some point, and plug this temperature in for y(t). This would give us the equation:

120 = Ae^(kt) + T sub a

Using algebra, we can solve for the constant A:

A = (120 - T sub a) / e^(kt)

Now, we can plug this value for A into our first equation and solve for the cooling rate constant k:

160 = [(120 - T sub a) / e^(kt)]*e^k + T sub a

Solving for k, we get k = ln(160 - T sub a) - ln(120 - T sub a).

Finally, we can plug this value for k into our equation for y(t) to solve for the temperature of the porridge when it reaches 120 degrees F:

y(t) = [(120 - T sub a) / e^([ln(160 - T sub a) - ln(120 - T sub a)]*t)] + T sub a

This would give us the temperature of the porridge at any given time t, including when it reaches 120 degrees F. I hope this explanation helps you understand the concept of Newton's Law of Cooling and how it applies to the given scenario.