Refridgerator Problem: Where Did I Screw Up?

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SUMMARY

The discussion centers on calculating the time required for a 2.90 kW space heater to generate the same amount of heat as a refrigerator with a coefficient of performance (COP) of 2.58 when freezing 1.45 kg of water at 19.4°C into ice at 0°C. The initial calculations yielded approximately 745 seconds, but this was incorrect due to a misunderstanding of the relationship between heat delivered and work done. The correct approach involves dividing the total heat delivered by the refrigerator (Q_h) by the COP to find the work (W) needed, leading to a revised formula: W = Q_c / COP.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the concept of Coefficient of Performance (COP).
  • Familiarity with heat transfer equations, including Q = mc(deltaT) and Q = mL_f.
  • Basic knowledge of power calculations, particularly P = E/t.
  • Ability to manipulate algebraic equations to solve for unknown variables.
NEXT STEPS
  • Review the principles of thermodynamics, focusing on Coefficient of Performance (COP) calculations.
  • Study heat transfer equations in detail, particularly the implications of latent heat (L_f) and specific heat (c).
  • Learn how to derive and manipulate equations related to power and energy in thermodynamic systems.
  • Explore practical applications of these concepts in refrigeration and heating systems.
USEFUL FOR

This discussion is beneficial for students and professionals in physics, engineering, and thermodynamics, particularly those involved in energy systems, refrigeration technology, and heat transfer analysis.

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Homework Statement



How long would a 2.90 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 2.58) does when it freezes 1.45 kg of water at 19.4°C into ice at 0°C?

Homework Equations


COP=Q_c/W
Q_h=Q_c + W
Q=cm(deltaT)
Q=mL_f
P=E/t

The Attempt at a Solution


I get something around 745s each time I do it, but that is wrong. The way I have been doing it is to find out the amount of energy it takes to change the water to ice which would be the work done by the refrigerator. Therefore Qc for the fridge would be coefficient of performance times work. Add Qc to work to get the amount of heat that exits the fridge and then divide by the number of watts used by the space heater to get the amount of time it would need to be run. Where did I go wrong in this reasoning? Thank you.

COP=Q_c/W therefore Q_c=COP*W
Q_total=mc(deltaT) + mL_f
Q_h=Q_c+W
Q_h=(COP+1)*(mc(deltaT) + mL_f)
Q_h/T=Power
[(COP+1)*(mc(deltaT) + mL_f)]/Power=T
[(2.58+1)*((1.45)(4186)(19.4) + (1.45)(33.5e4))]/2.9e3=745s
 
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EstimatedEyes said:

The Attempt at a Solution


I get something around 745s each time I do it, but that is wrong. The way I have been doing it is to find out the amount of energy it takes to change the water to ice which would be the work done by the refrigerator. Therefore Qc for the fridge would be coefficient of performance times work. Add Qc to work to get the amount of heat that exits the fridge and then divide by the number of watts used by the space heater to get the amount of time it would need to be run. Where did I go wrong in this reasoning? Thank you.

COP=Q_c/W therefore Q_c=COP*W
Q_total=mc(deltaT) + mL_f
Q_h=Q_c+W
Q_h=(COP+1)*(mc(deltaT) + mL_f)
Q_h/T=Power
[(COP+1)*(mc(deltaT) + mL_f)]/Power=T
[(2.58+1)*((1.45)(4186)(19.4) + (1.45)(33.5e4))]/2.9e3=745s
You have to divide your expression for Qh by COP:

The total heat delivered by the refrigerator is Qh = Qc + W = W(COP+1)

W = Qc/COP. That is what you have to work out by substituting:

Q_c = m(c\Delta T + L_f)

AM
 

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