# Differential Equations of Cooling Laws

1. Feb 13, 2012

### Biosyn

1. The problem statement, all variables and given/known data
When you turn on an electric heater, such as "burner" on a stove, its temperature increases rapidly at first, then more slowly, and finally approaches a constant high temperature. As the burner warms up, heat supplied by the electricity goes to two places.
i.) Storage in the heater materials, thus warming the heater
ii.) Losses to the room
Assume that heat is being supplied at a constant rate, R. The rate at which heat is stored is directly proportional to the rate of change of temperature. Let T be the number of degrees above room temperature. Let t be time in seconds. Then the storage rate is C(dT/dt). The proportionality constant, C (calories per degree), is called the heat capacity of the heater materials. According to Newton's law of cooling, the rate at which heat is lost to the room is directly proportional to T. The (positive) proportionality constant, h, is called the heat transfer coefficient.

a.) The rate at which heat is supplied to the heater is equal to the sum of the storage rate and the loss rate. Write a differential equation that expresses this fact.
b.) Separate the variables and integrate the differential equation. Explain why the absolute value sign is not necessary in this case. Transform the answer so that temperature, T, is in terms of time, t. Use the initial condition that T=0 when t=0.
c.) Suppose that heat is supplied at a rate, R=50 cal/sec. Assume that the heat capacity is C=2 cal/deg, and that the heat transfer coefficient is h=.04 (cal/sec)/deg. Substitute these values to get T in terms of t alone.

2. Relevant equations

dT/dt = k(M - T)

3. The attempt at a solution

Part a)

R = C (DT/dt) + hT

Part b)

R - hT = C(dT/dt)
→(R - hT) dt = C(dT)
→ ∫(R-hT) dt = ∫ C dT

Please tell me if I did it correctly, and help me integrate.

Thanks!

Same problem, but topic was locked:

2. Feb 13, 2012

### lanedance

R = C (dT/dt) + hT

should be
R = C (dT/dt) + h(T-Ts)
where Ts is the surrounding temperature, it does say we can assume T=0 though

now when you integrate, you attempt to separate variables, this essentially means you try and get all the T's on one side and all the t's on the other
$$R-hT= C \frac{dT}{dt}$$

This is complicated by the R term, so to seperatre and inetgrate, I would do the follwing transform first, let
$$U = hT-R$$
then
$$\frac{dT}{dt}=\frac{1}{h}\frac{dU}{dt}$$

now the equation is separable and it should be simpler to integrate, then you can transform back to T at the end
$$-U= \frac{C}{h} \frac{dU}{dt}$$