Solving Heat Pump Problem: 1/W = 1/(1-5/20)

AI Thread Summary
To determine the work per joule leakage for maintaining a house temperature of 20 degrees Celsius with an underground temperature of 5 degrees Celsius, the relevant equation is Q/W = 1/(1-(Tc/Th)). The initial attempt incorrectly used Celsius instead of absolute temperatures, leading to a miscalculation of work. The correct approach involves converting the temperatures to Kelvin, which resolves the discrepancy in the expected answer. Ultimately, the correct work required is approximately 0.05 Joules, highlighting the importance of using absolute temperature in thermodynamic calculations. Understanding this concept is crucial for accurately solving heat pump problems.
phoenix133231
Messages
7
Reaction score
0

Homework Statement


If the temperature underground is 5 degrees Celsius and you'd like to keep the house at 20 degrees Celsius, how much work per joule leakage must be done to maintain the house at that temperature?


Homework Equations


Q/W = 1/(1-(Tc/Th))


The Attempt at a Solution


I know this may be wrong... but since I needed to know the work "per joule" leakage, I assumed that Q = 1 Joule. Then, I plugged in the terms into the formula provided and ended up with the following:

1/W = 1/(1-5/20)

which gave me W = 4/3 Joules. I checked the back of the book and the answer ended up being 0.05 Joules.

Could someone please provide me some understanding to this problem, as well as a hint on how to start? Is the formula provided even useful?
 
Physics news on Phys.org
phoenix133231 said:

Homework Statement


If the temperature underground is 5 degrees Celsius and you'd like to keep the house at 20 degrees Celsius, how much work per joule leakage must be done to maintain the house at that temperature?


Homework Equations


Q/W = 1/(1-(Tc/Th))


The Attempt at a Solution


I know this may be wrong... but since I needed to know the work "per joule" leakage, I assumed that Q = 1 Joule. Then, I plugged in the terms into the formula provided and ended up with the following:

1/W = 1/(1-5/20)

which gave me W = 4/3 Joules. I checked the back of the book and the answer ended up being 0.05 Joules.

Could someone please provide me some understanding to this problem, as well as a hint on how to start? Is the formula provided even useful?

Consider if you need to use absolute temperatures - degK.

ice
 
LOL. I'm such a noob. That was exactly the problem. Thanks a lot!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
I was thinking using 2 purple mattress samples, and taping them together, I do want other ideas though, the main guidelines are; Must have a volume LESS than 1600 cubic centimeters, and CAN'T exceed 25 cm in ANY direction. Must be LESS than 1 kg. NO parachutes. NO glue or Tape can touch the egg. MUST be able to take egg out in less than 1 minute. Grade A large eggs will be used.
Back
Top