Why Is My Solution to Newton's Law of Cooling Equation Incorrect?

In summary, the equation dT/dt = -k(T - T_m) represents a cooling body where T is the temperature of the body, T_m is the temperature of the surroundings, -k is a constant, and t is time. Using the first order linear ODE integrating factor method, the solution is found to be T(t) = T_m + e^{-kt}(T_0 - T_m), where T_0 is the initial temperature and T(t) is the temperature at time t. This solution can also be obtained by using separation of variables, which is less prone to error.
  • #1
animboy
27
0

Homework Statement



dT/dt = -k(T - T_m)

T is the temperature of the body,

T_m is the temperature of the surroundings,

-k is some contant

and t is ofcourse time

Homework Equations



no idea

The Attempt at a Solution



I tried solving this using first order linear ODE integrating factor method:

so in standard form it can be written as-

T' + kT = T_m

let p(t) = k
then u(t) = exp(∫ k dt)

u(t) = exp(kt) we can forget about the constant.

so multiplying ODE throughout by u(x) :

exp(kt)*(T' + kT) = exp(kt)*(T_m)

integrate both sides

exp(kt)*T = [exp(kt)*(T_m)]/k

divide both sides by u(t)

T = (T_m)/k

my book says this is wrong, why?
 
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  • #2
animboy said:

Homework Statement



dT/dt = -k(T - T_m)

T is the temperature of the body,

T_m is the temperature of the surroundings,

-k is some contant

and t is ofcourse time

Homework Equations



no idea

The Attempt at a Solution



I tried solving this using first order linear ODE integrating factor method:

so in standard form it can be written as-

T' + kT = T_m

let p(t) = k
then u(t) = exp(∫ k dt)

u(t) = exp(kt) we can forget about the constant.

so multiplying ODE throughout by u(x) :

exp(kt)*(T' + kT) = exp(kt)*(T_m)

integrate both sides

exp(kt)*T = [exp(kt)*(T_m)]/k

divide both sides by u(t)

T = (T_m)/k

my book says this is wrong, why?

A few issues:

Going from here: dT/dt = -k(T - T_m)

to here: T' + kT = T_m

you missed out a factor of k on the RHS.

Your equation should be T' + kT = kT_m.

Your Integrating Factor is OK (although this can be simply solved by separation of variables). But when you do the integration, you didn't put in bounds, which is a very important step when solving a physical problem.

Both sides are being integrated from t = 0 to t (the latter just represents a general time 't').

The temperature should be represented by T_0 initially, and T(t) at time t.

Evaluate both sides as definite integrals with the correct bounds and expressions for T, and you should get this result:

[itex]T(t) = T_m + e^{-kt}(T_0 - T_m)[/itex]

from which you can easily deduce the physical behaviour of the cooling body and see that it matches up to intuition, starting at T_0 and ending at infinite time at T_m.

As I mentioned, you don't need IF to solve this, you can just use separation of variables; it's less prone to error. Try it.
 

FAQ: Why Is My Solution to Newton's Law of Cooling Equation Incorrect?

What is Newton's law of cooling?

Newton's law of cooling is a mathematical representation of how the temperature of a body changes over time when it is in contact with a cooler or warmer surrounding environment. It states that the rate of change of temperature of an object is proportional to the difference between its own temperature and the ambient temperature.

How is Newton's law of cooling expressed in terms of an Ordinary Differential Equation (ODE)?

The ODE representation of Newton's law of cooling is dT/dt = -k(T-Ta), where T is the temperature of the object, Ta is the ambient temperature, k is a constant that depends on the properties of the object and its surroundings, and dT/dt is the rate of change of temperature over time.

How is this ODE solved?

The solution to this ODE depends on the initial conditions (the temperature at time t=0) and the value of the constant k. The general solution is T(t) = Ta + Ce-kt, where C is the constant of integration. To solve for C, we need to have a value for T(t) at a specific time, which can be obtained from the initial conditions.

How does the value of the constant k affect the solution?

The value of k determines the rate at which the temperature of the object changes. A higher value of k means that the object cools down or heats up faster, while a lower value of k means a slower rate of change. This constant is influenced by factors such as the material and size of the object, as well as the properties of the surrounding environment.

Can Newton's law of cooling be applied to real-life situations?

Yes, Newton's law of cooling can be applied to various real-life situations involving heat transfer, such as the cooling of a cup of hot coffee, the heating of a room with a heater, or the cooling of a human body when exposed to a cold environment. However, it is important to note that this law is an idealization and may not perfectly describe the behavior of all objects in all situations.

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