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Refrigerator Question

  • Thread starter xmonsterx
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  • #1
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Homework Statement



PROBLEM:
Refrigeration units can be rated in "tons". A 1-ton air conditioning system can remove sufficient energy to freeze 1 British ton (2000pounds) = 909 kg) of 0 degree C water into 0 degree C ice in one 24-h day.


QUESTION:
If, on a 40 degree C day, the interior of a house is maintained at 20 degree C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs $0.10 per kWh and that the unit's coefficient of performance is 15% that of an ideal refrigerator. 1 kWh= 3.60 times 10^6 J.

PLEASE GIVE THE ANSWER IN $/hr !!!!!!!!!!!!!!!!!!!!!!!!!!!


Homework Equations


COP = Q_L / W


The Attempt at a Solution


COP = (40 +273) / W ???

i dont even know where to start!! please help i need this badlyyy.
 

Answers and Replies

  • #2
Andrew Mason
Science Advisor
Homework Helper
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Homework Statement



PROBLEM:
Refrigeration units can be rated in "tons". A 1-ton air conditioning system can remove sufficient energy to freeze 1 British ton (2000pounds) = 909 kg) of 0 degree C water into 0 degree C ice in one 24-h day.


QUESTION:
If, on a 40 degree C day, the interior of a house is maintained at 20 degree C by the continuous operation of a 5-ton air conditioning system, how much does this cooling cost the homeowner per hour? Assume the work done by the refrigeration unit is powered by electricity that costs $0.10 per kWh and that the unit's coefficient of performance is 15% that of an ideal refrigerator. 1 kWh= 3.60 times 10^6 J.

PLEASE GIVE THE ANSWER IN $/hr !!!!!!!!!!!!!!!!!!!!!!!!!!!


Homework Equations


COP = Q_L / W


The Attempt at a Solution


COP = (40 +273) / W ???

i dont even know where to start!! please help i need this badlyyy.
First you have to find how much heat in joules that you have to remove in an hour. (Hint: how much heat does one have to remove from 909 kg of water at 0 C to turn it into ice at 0C?)

Then you have to determine the COP of the air conditioner to determine how much work must be performed to run it for one hour (in Joules and then KwH).

AM
 
  • #3
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First you have to find how much heat in joules that you have to remove in an hour. (Hint: how much heat does one have to remove from 909 kg of water at 0 C to turn it into ice at 0C?)

Then you have to determine the COP of the air conditioner to determine how much work must be performed to run it for one hour (in Joules and then KwH).

AM
i dont know what formula to use to find how much heat is removed in an hour. I need additional help/ hints. thanks
 
  • #4
Andrew Mason
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i dont know what formula to use to find how much heat is removed in an hour. I need additional help/ hints. thanks
You need to use the heat of fusion of water, which s 333.55 kJ/kg.

AM
 
  • #5
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You need to use the heat of fusion of water, which s 333.55 kJ/kg.

AM
wait.. heat of fusion which is Lf = mL = 333.55 .... how does this help me solve for the answer?

what if i use deltaQ = mCdeltaT ???

delta Q = 909kg * 1cal/g C * ((40+273)-(20+273)) ??

does this look right? or am i totally off?
 
  • #6
Andrew Mason
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wait.. heat of fusion which is Lf = mL = 333.55 .... how does this help me solve for the answer?

what if i use deltaQ = mCdeltaT ???

delta Q = 909kg * 1cal/g C * ((40+273)-(20+273)) ??

does this look right? or am i totally off?
You cannot use [itex]\Delta Q = mC\Delta T[/itex] because there is no change in temperature. Only a change of state. So you have to use:

[tex]\Delta Q = mL_f[/tex] where [itex]L_f[/itex] is the heat of fusion of water.

AM
 
  • #7
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You cannot use [itex]\Delta Q = mC\Delta T[/itex] because there is no change in temperature. Only a change of state. So you have to use:

[tex]\Delta Q = mL_f[/tex] where [itex]L_f[/itex] is the heat of fusion of water.

AM
ok so...
delta Q = (909kg) * (333.55) = 3.0*10^5

so then do i plug this number in this equation --> COP= Q_L / W --> COP = Q_L / Q_H -Q_L ??

am i on the correct path??
 
  • #8
Andrew Mason
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ok so...
delta Q = (909kg) * (333.55) = 3.0*10^5

so then do i plug this number in this equation --> COP= Q_L / W --> COP = Q_L / Q_H -Q_L ??

am i on the correct path??
You have to find the COP of this refrigerator. Do you know how to determine the COP of a Carnot refrigerator operating between 20C (Tc) and 40C (Th)? (COP = Qc/W). Take 15% of that. (I get 2.2 for the COP of this machine).

AM
 

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