What is the Proof for Newton's Law of Cooling Formula Using Quadratic Equation?

[guyvsdcsniper]

Homework Statement

A building is maintained at temperature
TH with a reversible heat pump operating between the building and a
colder environment at temperature TC < TH. The heat pump consumes electrical
power at a constant rate W . The building also loses heat according to
Newton’s law of cooling, that is, at a rate (TH – TC) where is constant.
Show that the building temperature is maintained at temperature

Relevant Equations

dQrev/T = 0

I have went about this problem many different ways but cannot seem to come up with the answer. I am essentially trying to prove the formula provided in the ss of the problem.Could someone help me and tell me if I am approaching this wrong?

 

[collinsmark]

Homework Statement:: A building is maintained at temperature
TH with a reversible heat pump operating between the building and a
colder environment at temperature TC < TH. The heat pump consumes electrical
power at a constant rate W . The building also loses heat according to
Newton’s law of cooling, that is, at a rate (TH – TC) where is constant.
Show that the building temperature is maintained at temperature
Relevant Equations:: dQrev/T = 0

I have went about this problem many different ways but cannot seem to come up with the answer. I am essentially trying to prove the formula provided in the ss of the problem.Could someone help me and tell me if I am approaching this wrong?

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View attachment 297168

I haven't gone through your entire approach (my eyes aren't what they used to be). But since the heatpump is reversible, we can assume it operates as an ideal Carnot engine. It has the efficiency:

\eta = \frac{\dot{W}}{\dot{Q_H}} = 1 - \frac{T_C}{T_H}

Now, what's \dot{Q_H}? It's the rate at which heat leaves the hot-sink, right? The rate that heat leaves the building. You have another formula for that rate.

I would start there. (Hint: There's some signifiant algebra involved, and if you find yourself with a quadratic, you might be on the right track.)

 

[Steve4Physics]

I have went about this problem many different ways but cannot seem to come up with the answer. I am essentially trying to prove the formula provided in the ss of the problem.Could someone help me and tell me if I am approaching this wrong?

View attachment 297169
View attachment 297168

Not my area but, since no one has replied yet, maybe this will help. I think a different approach is required,

Do you know how to express the coefficient of performance (heating) of the heat pump in terms of $T_C$ and $T_H$?

If not, take a look here: https://en.wikipedia.org/wiki/Coefficient_of_performance

Energy is being supplied to the building at a rate of $\dot W .COP_{heating}$.

Since this will equal the rate of heat loss, $α(T_H-T_C)$, you can write an equation which should give a quadratic equation in $T_H$ that you can solve.

But the algebra looks like it will be a bit messy!

EDIT. Aha! @collinsmark beat me to it!

 

[guyvsdcsniper]

Not my area but, since no one has replied yet, maybe this will help. I think a different approach is required,

Do you know how to express the coefficient of performance (heating) of the heat pump in terms of $T_C$ and $T_H$?

If not, take a look here: https://en.wikipedia.org/wiki/Coefficient_of_performance

Energy is being supplied to the building at a rate of $\dot W .COP_{heating}$.

Since this will equal the rate of heat loss, $α(T_H-T_C)$, you can write an equation which should give a quadratic equation in $T_H$ that you can solve.

But the algebra looks like it will be a bit messy!

EDIT. Aha! @collinsmark beat me to it!

I just checked my book and it does mention COP but it only mentions the COP of a refrigerator. Thank you for the wiki page. I will try it using this new approach.

 

[guyvsdcsniper]

I haven't gone through your entire approach (my eyes aren't what they used to be). But since the heatpump is reversible, we can assume it operates as an ideal Carnot engine. It has the efficiency:

\eta = \frac{\dot{W}}{\dot{Q_H}} = 1 - \frac{T_C}{T_H}

Now, what's \dot{Q_H}? It's the rate at which heat leaves the hot-sink, right? The rate that heat leaves the building. You have another formula for that rate.

I would start there. (Hint: There's some signifiant algebra involved, and if you find yourself with a quadratic, you might be on the right track.)

Hey, so I got close to the solution of this problem but kept coming up short. I ended up referring to a solution someone else posted but am I bit lost on the algebra.

I am a bit lost on the jump to forming a quadratic equation. How is it that T_h² turns into T_h.

 

[Steve4Physics]

Hey, so I got close to the solution of this problem but kept coming up short. I ended up referring to a solution someone else posted but am I bit lost on the algebra.

I am a bit lost on the jump to forming a quadratic equation. How is it that T_h² turns into T_h.

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If you are OK up to:
$T_H^2 - T_H(2 T_C + \frac {\dot W}{\alpha}) + T_C^2 = 0$
then compare this to the standard quadratic equation
$ax^2+ bx+c=0$.

Instead of $x$ we are using $T_H$.
The constants are:
$a = 1$
$b = -(2T_C + \frac {\dot W}{\alpha})$ and
$c = T_C^2$

To get the value of $x$ (i.e. $T_H$) we use $x = \frac{ -b± \sqrt {b^2 – 4ac)}} {2a}$.

 

[guyvsdcsniper]

If you are OK up to:
$T_H^2 - T_H(2 T_C + \frac {\dot W}{\alpha}) + T_C^2 = 0$
then compare this to the standard quadratic equation
$ax^2+ bx+c=0$.

Instead of $x$ we are using $T_H$.
The constants are:
$a = 1$
$b = -(2T_C + \frac {\dot W}{\alpha})$ and
$c = T_C^2$

To get the value of $x$ (i.e. $T_H$) we use $x = \frac{ -b± \sqrt {b^2 – 4ac)}} {2a}$.

That makes so much sense. Thank you so much