Mathematics: 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?

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Newton s law of cooling is: du/dt= -k(u-t) where

leah1386

16.01.2022

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

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

$\frac{du}{dt}= -k (u-T)$

We can reorder the expression like this:

$\frac{du}{u-T} = -k dt$

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

$\frac{dw}{w} =-k dt$

IF we integrate both sides we got:

ln |w| = -kt +C

If we apply exponential in both sides we got:

$w = e^{-kt} *e^c$

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

$u(t)= T + C_1 e^{-kt}$

We can also express the solution in the following terms:

$u(t) = (T_i -T_{amb}) e^{kt} +T_{amb}$

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

$u(t) = (70-11) e^{-0.15t} +11$

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

$32 = 59 e^{-0.15 t} +11$

$21=59 e^{-0.15 t}$

$\frac{21}{59} = e^{-0.15 t}$

If we apply natural logs on both sides we got:

$ln (\frac{21}{59}) =-0.15 t$

$t = \frac{ln(\frac{21}{59})}{-0.15}=6.887 hr$

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

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