Heating Your Kitchen with Your Refrigerator

Click For Summary
SUMMARY

The discussion centers on calculating the time required for a 3.50 kW space heater to generate the same amount of heat as a refrigerator with a coefficient of performance (COP) of 3.13, which freezes 1.46 kg of water from 21.4°C to 0°C. The calculations yield a total heat transfer of 618,427 J, leading to a work output of 197,581 J. Dividing this work by the heater's power results in a time of 56.45 seconds. However, the solution is incorrect due to a misunderstanding of where the heat is dissipated during the process.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat transfer principles.
  • Familiarity with the concept of Coefficient of Performance (COP).
  • Knowledge of basic physics equations related to work and energy.
  • Ability to perform calculations involving power, energy, and time.
NEXT STEPS
  • Review the principles of thermodynamics, focusing on heat transfer and energy conservation.
  • Study the Coefficient of Performance (COP) in detail, including its implications in heating and cooling systems.
  • Learn how to apply the first law of thermodynamics to practical problems involving heat and work.
  • Explore practical applications of space heaters and refrigerators in energy efficiency contexts.
USEFUL FOR

Students in physics or engineering, HVAC professionals, and anyone interested in thermodynamic systems and energy efficiency in heating and cooling applications.

2001
Messages
1
Reaction score
0

Homework Statement



How long would a 3.50 kW space heater have to run to put into a kitchen the same amount of heat as a refrigerator (coefficient of performance = 3.13) does when it freezes 1.46 kg of water at 21.4°C into ice at 0°C?

m=1.46 kg
c_water=4186 J/kg
Delta T= 21.4°C
L_fusion= 3.34x10^5 J/kg
COP= 3.13
P= 3.50 kW

Homework Equations



Q_c = m x c_water x Delta T + m x L_f

COP = Q_c/W

P = W/t

The Attempt at a Solution



I plugged everything into the equation for the heat and got 618427 J. Then I divided that by the coefficient of performance to get that work = 197581 J. Then, I divided that by 3500 W for the power and found that t = 56.45 s. And apparently that's not the answer. I'm not sure where I went wrong!
 
Physics news on Phys.org
You computed the work done to transfer an amount of heat. Where did that heat go?
 

Similar threads

First Law of Thermo: Audience body heat warming a concert hall
  • · Replies 25 ·
Replies
25
Views
2K
Optimizing Refrigerator Efficiency: Solving for Necessary Power Draw
  • · Replies 2 ·
Replies
2
Views
4K
I try to solve a thermodynamics problem on heat transfer
  • · Replies 3 ·
Replies
3
Views
2K
Solving Heat Engine & Refrigerator Problems
  • · Replies 2 ·
Replies
2
Views
3K
Why use c_p and not c_v as specific heat - Thermodynamics
  • · Replies 2 ·
Replies
2
Views
2K
High School Heat pump as a heat engine that operates in reverse
  • · Replies 19 ·
Replies
19
Views
5K
Physics Homework about using Electricity to Heat Flowing Water
  • · Replies 9 ·
Replies
9
Views
2K
Mass of Ice required to achieve a certain final Temperature
  • · Replies 3 ·
Replies
3
Views
2K
Thermodynamics - Hypothermia - Heat loss
  • · Replies 12 ·
Replies
12
Views
2K
Refridgerator Problem: Where Did I Screw Up?
  • · Replies 1 ·
Replies
1
Views
3K