contestada

Newton's law of cooling is:

du/dt= -k(u-T)

where u(t) is temperature of an object, t is in hours, T is a constant ambient temperature, and k is a positive constant.

Suppose a building loses heat in accordance with Newton's law of cooling. Suppose that the rate constant k has the value 0.15 hr ? 1 . Assume that the interior temperature is Ti = 70?F, when the heating system fails. If the external temperature is T = 11? F, how long will it take for the interior temperature to fall to T1 = 32?F?

Respuesta :

Answer:

[tex] t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr[/tex]

So it would takes approximately 6.9 hours to reach 32 F.

Step-by-step explanation:

For this case we have the following differential equationÑ

[tex]\frac{du}{dt}= -k (u-T)[/tex]

We can reorder the expression like this:

[tex] \frac{du}{u-T} = -k dt[/tex]

We can use the substitution [tex] w = u-T[/tex] and [tex] dw =du[/tex] so then we have:

[tex] \frac{dw}{w} =-k dt[/tex]

IF we integrate both sides we got:

[tex] ln |w| = -kt +C[/tex]

If we apply exponential in both sides we got:

[tex] w = e^{-kt} *e^c[/tex]

And if we replace w = u-T we got:

[tex] u(t)= T + C_1 e^{-kt}[/tex]

We can also express the solution in the following terms:

[tex] u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}[/tex]

For this case we know that [tex] k =-0.15 hr[/tex] since w ehave a cooloing, [tex] T_{i}= 70 F, T_{amb}=11F[/tex], we have this model:

[tex] u(t) = (70-11) e^{-0.15t} +11[/tex]

And if we want that the temperature would be 32F we can solve for t like this:

[tex] 32 = 59 e^{-0.15 t} +11[/tex]

[tex] 21=59 e^{-0.15 t}[/tex]

[tex] \frac{21}{59} = e^{-0.15 t}[/tex]

If we apply natural logs on both sides we got:

[tex] ln (\frac{21}{59}) =-0.15 t[/tex]

[tex] t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr[/tex]

So it would takes approximately 6.9 hours to reach 32 F.