# Writing a differential equation to describe cooling

1. Jun 24, 2012

### 1MileCrash

1. The problem statement, all variables and given/known data

Newton's Law of Cooling states that the temperature of an object changes at a rate proportional to the difference between the temperature of the object itself and the temperature of its environment.

Suppose the ambient air temperature is 70*F and that the rate constant is 0.05min^-1.

Write a differential equation for the temperature change the object undergoes.

2. Relevant equations

3. The attempt at a solution

Just starting with DE in preparation for fall.

I just wrote:

let q be heat, t be time in minutes.

(dq/dt) = 0.05(q-70)

...

Is that really it?

And solving this differential equation, would mean finding a function, q(t), such that its derivative is equal to itself, minus 70, times 0.05, for any value of t?

Does that mean that modeling and solving differential equations is mainly a method of finding an equation that models a situation by examining the behavior of its rate of change?

2. Jun 24, 2012

### Dick

Yes, it is that easy, if your 'heat' variable q means temperature.

3. Jun 25, 2012

### HallsofIvy

Staff Emeritus
Actually, it is NOT quite that easy! If the initial temperature, q, is greater than the ambient temperature, 70, then the object will cool! That is, q will decrease and dq/dt is negative. If the initial temperature, q, is lower than the ambient temperature, 70, then the object will warm! That is, q will increase and dq/dt is positive.

dq/dt= 0.05(70- q) or dq/dt= -0.05(q- 70)

4. Jun 25, 2012

### 1MileCrash

Since it introduced it as the "law of cooling" my instinct was to write a de for the rate of cooling, not heating.

5. Jun 25, 2012

### Dick

Halls is right. I missed the sign. You have to think about the physics of the situation to get the right sign. If u is greater than 70 it should be cooling, if u is less the 70 it should be heating.

6. Jun 25, 2012

### 1MileCrash

I understand what you are saying. I mean that I "chose" a drop in temperature to be positive. I see why this may be considered awkward.

7. Jun 25, 2012

### Dick

It's worse than awkward. It's wrong. I was sloppy in overlooking it. Sorry.

8. Jun 25, 2012

### 1MileCrash

Still easy. :P