differential equation news law of heating with furnace?

Let T be the temperature in °F of the lecture hall.

The rate of change due to heat loss to outside is

- K∙(T - T_outside) = - 0.5 hr⁻¹∙(T - 40)

The rat of change due to heat absorption from the furnace is:

+ K₁∙(T_furnace - T) = + 2 hr⁻¹∙(70 - T)

The overall rate of change is the sum of the two:

Assuming t is measured in hr you can omit the unit hr⁻¹ for K and K₁,. The you get

dT/dt = + 2∙(70 - T) - 0.5∙(T - 40) = 160 - 2.5∙T

<=>

dT/dt = 2.5∙(64 - T)

Separation of variables leads to

-1/(64 - T) dT = -2.5 dt

=>

∫ -1/(64 - T) dT = ∫ -2.5 dt

<=>

ln(64 - T) = - 2.5∙t + c

(C is the constant of integration)

T = 64 - e^(- 2.5∙t + c) = 64 - e^(c)∙e^(- 2.5∙t) = 64 - C∙e^(- 2.5∙t)

with C = e^(c)

Apply initial condition:

T(0) = 40

<=>

64 - C∙e^(- 2.5∙0) = 40

<=>

64 - C = 40

<=>

C = 24

Hence temperature of the lecture hall in °F t hours after 7:00AM is given by:

T(t) = 64 - 24∙e^(- 2.5∙t)

What will bethe temperature inside the lecture hall at 8:00 AM?

t = 1

=>

T = 64 - 24∙e^(- 2.5∙t) = 62

When will the temperature inside the lecture hall reach 65 °F?

Never

Consider the equation for T. As time t approaches infinity the exponential term vanishes, i.ie

lim|t→∞| { 64 - 24∙e^(- 2.5∙t)) | = 64

That means the temperature of the hall will never exceeds 64°F.

The physical meaning of this result is, that at this temperature the rate of heat absorption from furnace and rate of heat loss to the outside become equal, thus:

dT/dt = 0 for T = 64

So no further change of T takes place.

Schmiso's response is almost right.

Except he forgot to solve for the K of the heating system. Let's call it Ku.

1/K = 2, and 1/K1 (the time constant for the building + heating system) = 1/2

Therefore, K = 1/2, and K+Ku = 2. So, Ku = 2 - K = 2 - 1/2 = 3/2

dT/dt = K(Toutside-T)+Ku(Tfurnace-T)

Thus your differential equation is:

dT/dt =1/2(40-T)+3/2(70-T)

Or, dT/dt=125-2T

Solve using separation or linear techniques, (linear ends up being easier imo) and:

T(t)=125/2+Ce^(-2t)

Plug in the initial condition T(0)=40

40=125/2+C -> C = -45/2

So, T(t)=125/2-45/2e^(-2t)

And finally, @8am t=1

T(1)=125/2-45/2e^(-2)=~59.5

and, for t when T=65,

65=125/2-45/2e^(-2t)

-1/9=e^(-2t)

t= -1/2ln(-1/9) ... this is an imaginary number, and thus the lecture hall will still never reach 65F

you can easily see what the max temp would be though... Go back to our differential equation:

dT/dt=125-2T

and see what the temp would be when the temp change with respect to time is zero:

dT/dt=0, so,

0=125-2T

2T=125

T=125/2=62.5

So as t->infinity, T->62.5

This means, given an infinite amount of time, the temp would stabilize at 62.5F