First order seperable ode with IC's

[wtmoore]

Homework Statement

Question

According to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given byAccording to Newton’s Law of Cooling, the rate at which a substance cools in air is proportional to the difference between the temperature of the substance and that of air. The differential equation is given by:

dT/dt=-k(T-T​0)

where T is the temperature of the substance, T0 is the temperature of air, k is a constant of proportionality and t is the time. Find the general solution of this equation.
If T0=30°C then find the solution subject to the initial condition that T=100°C at time t=0.
Suppose an experiment is carried to determine the physical constant k. If the substance cools from 100°C to 70°C in 15 minutes then find k. Hence find the time taken for the temperature to fall from 100°C to 40°C

The Attempt at a Solution

Ok guys, Just so you know where I'm at, I can do seperable ODES quite easily, however, I am useless and wordy questions and I don't know how to set it up with IC's in. I need to bring the T under the dT, but don't know how to separate it from -k(T-T0).

If I multiply out I get

dT/dt=-kT+T0

and I know know how to get it under the dT, can someone explain how I do this? I may need further help with the initial conditions but I need to know how to find the general solution for this type of ode with IC's first.

 

[Dick]

Writing it in the form dT/(T-T0)=(-k)dt is separated enough. T0 and k are just constants. You can integrate both sides. Use a trivial change of variables on the dT/(T-T0) side.

 

[wtmoore]

Ok,

so now the general solution is:

T=Cexp(-kt)+T0

particular solution is

T=70exp(-kt)+30 when T=100 when t=0 and T0=30

However for the next bit I don't understand,

Am I doing T(15)=70?

Does this give

T=Cexp(-15k)+30?

So 70=70exp(-15k)+30

k=0.0373

Then T=40

40=70exp(-0.0373t)+30

t=52mins?

 

[Dick]

Ok,

so now the general solution is:

T=Cexp(-kt)+T0

particular solution is

T=70exp(-kt)+30 when T=100 when t=0 and T0=30

However for the next bit I don't understand,

Am I doing T(15)=70?

Does this give

T=Cexp(-15k)+30?

So 70=70exp(-15k)+30

k=0.0373

Then T=40

40=70exp(-0.0373t)+30

t=52mins?

Looks ok to me.

 

[wtmoore]

thanks dick