Physics Help Forum

Physics Help Forum Feed Site Feed

  #1  
Old 05-31-2009, 05:39 PM
Junior Member
 
Join Date: May 2009
Posts: 1
Thanks: 1
Thanked 0 Times in 0 Posts
ricez375 is on a distinguished road
Default refrigerator problem

A freezer is rated at 1100 watts. 90% of its power is used in the freezing process. How long will it take to make 3kg of ice at -10°C from water at 25°C?

please explain in detail, im having trouble understanding heat.
Reply With Quote
Advertisement
 
  #2  
Old 05-31-2009, 11:29 PM
PHF Helper
 
Join Date: Feb 2009
Posts: 1,274
Thanks: 221
Thanked 444 Times in 398 Posts
physicsquest is a glorious beacon of lightphysicsquest is a glorious beacon of lightphysicsquest is a glorious beacon of lightphysicsquest is a glorious beacon of lightphysicsquest is a glorious beacon of light
Default

A freezer is rated at 1100 watts. 90% of its power is used in the freezing process. How long will it take to make 3kg of ice at -10°C from water at 25°C?

3 kg of ice is equivalent to 3 kg of water because mass is conserved.

The freezer takes away heat by doing work.

So heat is taken away from 3 kg of water at 25 degrees to get 3 kg of water at 0 deg.
Then further heat is taken away from 3 kg of water to convert it to 3 kg of ice.
Finally heat is further taken away from 3 kg of ice at zero deg to get 3 kg of ice at -10 deg.

It is easier to understand if we reverse this problem, i.e. supply heat to 3 kg ice at -10 deg to convert it to 3 kg water at 25 deg using a heater.
The heat required for this process will be the same as the heat taken away in the previous case by the freezer.

The heat required will be

Q = 3 x Ci x [0 - (-10)] + (3 x Li ) + 3 x Cw x (25-0) where

Ci = Sp heat of ice, Li = Latent heat of fusion of ice, and
Cw = Sp heat of water. This result will be in calories. Convert to J.

The heater supplies (freezer takes away) Q J of energy at efficiency of 90% . Let E be the energy supplied to to the heater to get Q J of energy. Since the efficiency = 90%, we have Q/E = 0.9 or E = Q/0.9.

Now Power = energy / time So, Energy = Power x time So

E = 1100 x t. Solve for t.

Last edited by physicsquest; 06-01-2009 at 04:49 AM.
Reply With Quote
The Following User Says Thank You to physicsquest For This Useful Post:
ricez375 (06-01-2009)

Donate to PHF
  #3  
Old 07-01-2009, 12:56 PM
Junior Member
 
Join Date: Jun 2009
Posts: 24
Thanks: 0
Thanked 2 Times in 2 Posts
george is on a distinguished road
Default

You have not taken into account the coefficient of performance.
The coefficient of performance COP of a heat pump is the ratio between the power of the heating and electric power absorbed from the network. Spread depends on the temperature of cold source and heating source.
Reply With Quote
  #4  
Old 07-03-2009, 02:44 PM
Junior Member
 
Join Date: Jul 2009
Posts: 2
Thanks: 0
Thanked 0 Times in 0 Posts
AniMuS is on a distinguished road
Default

So except for the Heat Q on the right, this is what we have:



[LaTeX Error: Can't write to directory]

[LaTeX Error: Can't write to directory]?

The COP of the refrigerator is less or equal to [LaTeX Error: Can't write to directory]

[LaTeX Error: Can't write to directory] is ambient temp.
[LaTeX Error: Can't write to directory] is water temp.

Not sure if it's, just spitting out ideas.

Last edited by AniMuS; 07-04-2009 at 05:28 AM.
Reply With Quote
  #5  
Old 07-03-2009, 09:19 PM
Junior Member
 
Join Date: Jun 2009
Posts: 24
Thanks: 0
Thanked 2 Times in 2 Posts
george is on a distinguished road
Default

"Now, COP = 0.9 or 90%"-Fals
COP is always supraunitar(2;2,5;3...)
COP" ( coefficient of performance ) si "EER" ( energetic efficiency of refrigeration )
COP=QL/W -corect
.:Kelvin PDC - promovarea si implementarea sistemelor neconventionale de producere a energiei:. (see diagram)

Last edited by george; 07-04-2009 at 05:13 PM.
Reply With Quote
Reply

Thread Tools
Display Modes

Posting Rules
You may not post new threads
You may not post replies
You may not post attachments
You may not edit your posts

BB code is On
Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are Off
Pingbacks are Off
Refbacks are Off
Forum Jump


All times are GMT -7. The time now is 09:37 AM.


Powered by vBulletin® Version 3.7.3
Copyright ©2000 - 2010, Jelsoft Enterprises Ltd.
SEO by vBSEO 3.2.0 ©2008, Crawlability, Inc.
©2009 Physics Help Forum

Physics Help Forum is a community of physics forums with an emphasis on physics help in all levels of physics.
Register to post your physics questions on the message board.