Newton’s law of cooling, difficult question

By Newton’s law of cooling the temperature T of a cooling object as a function of time is

TO - initial temperature,

T1 is ambient temperature (meaning the temperature outside the cooling object)

k is constant

W

-32C was the te,perature when electricity was cut off and heating of a house stopped working. After 2 hours from the power cut the inside temperature was measured to be 19,2 C, and after that the following measurements were done:

2,5 hours: 18,3 oC

3,0 hours: 17,5 oC

3,5 hours: 16,6 oC

4,0 hours: 15,8 oC

4,5 hours: 14,9 oC

5,0 hours: 14,2 oC

Supposing Newton’s cooling law to be valid in this case,

Search:

k

T0 initial temperature (which is the temperature at moment t=0 when the power was cut off). Try also to predict when the inside temperature will be dropped below zero.

__________________________________________________ __________________________________________________ __

my attempt:

19.2-(-32)=(T0-32)e-k(2)

14.2-(-32)=(T0-32)e-k(5)

51.2 (T0-32)e-k(2)

46.2 (T0-32)e-k(5)

1.11 e2,5k

k= 0.041103894

k= ln(1.11)/2.5

Substitude k into the first equation

51.2=(T0-32)e-2k

51.2=(T0-32)e-2((ln(1,11)/2,5))

and now ?????????????

Re: Newton’s law of cooling, difficult question

I have no idea what you are **doing**. You posted this in "differential equations" but there is no differential equation. You title this "Newton's law of cooling" but never say what that is.

Re: Newton’s law of cooling, difficult question

Hi Plastique!

Newton's Law of Cooling is defined as:

dT/dt=k(T-Tm) where k is some constant and Tm is the ambient temperature.

Solving the ODE gives:

T=ce^(kt)+Tm

We can set Tm= -32

so, T=ce^(kt)-32

Using the initial conditions t=2,T=19.2, our equation becomes:

19.2=ce^(2k)-32 ---> 51.2=ce^(2k) ---> (51.2/c)=e^(2k) ---> ln(51.2/c)/2=k

We can now substitute k back into the equation:

T=ce^[ln(51.2/c)t/2]-32

Now we use the next initial condition (which can be any of the given values by the way) t=5,T=14.2, and our equation becomes:

14.2=ce^[ln(51.2/c)5/2]-32

And solving for c gives:

c= 54.83047 Remember that ln(51.2/c)5/2=ln((51.2/c)^(5/2))

We can now also find k by substituting c into our equation for k:

k=ln(51.2/c)/2 ---> k=ln(51.2/54.83047)/2= -0.03425

So we now have both c=54.82047 and k=-0.03425 which makes our equation:

T=(54.82047)e^(-0.03425t)-32

Finding the initial temperature of the house comes from setting t=0 which will give T0= 22.83047 (a reasonable temperature).

To find how long it takes for the temperature of the house to drop below 0 degrees Celsius, set T=0 and solve for t.

t=15.72 hours.