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Heat Exchange and Efficiency in a Fridge

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data
    In a 30% efficient, -16°C fridge, how much electrical energy will be used up to lower the temperature of 250 mL of 75°C water? If the fridge is rated at 150 W, how long will this process take?

    Specific Heat Capacity of Water = 4200 J/Kg×K
    Specific Heat Capacity of Ice = 2000 J/Kg×K
    Latent Heat of Fusion for Water = 333000 J/Kg

    2. Relevant equations
    Efficiency=EnergyOutput/EnergyInput × 100%

    3. The attempt at a solution
    I've found the QTotal (Q1, Q2 and QF), which is -3500J, but I'm not sure what to do with that info, or even what it means in regards to the question. The wording is also quite confusing, and I'm not sure whether used up means the input or the output. Any help would be greatly appreciated!!
  2. jcsd
  3. Apr 25, 2015 #2

    Simon Bridge

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    First you should describe what "that info" is - i.e. what does "Qtotal" mean?
    Do you know how much energy needs to be extracted from the water to cool it?

    Note: "used up" means exactly that - the fridge uses energy to make the water colder ... where does that energy go?
    (You should not be thinking in terms of "input" or "output" - that will confuse you ... just follow the flow of energy.)

    I think there is something wrong with the question but I want you to wrap your head around the concepts first before discussing it. It does not actually affect how you do the problem.
    Last edited: Apr 25, 2015
  4. Apr 25, 2015 #3


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    3500 J is way too low. That allows cooling the water by a ~3 Kelvin.

    How much (electric input) power would a 100% efficient fridge need to remove 1 J of heat from the -16°C interior if the outside is at room temperature?
  5. Apr 25, 2015 #4
    I'm not sure what you mean by "how much energy needs to be extracted from the water to cool it"... Would that be QTotal?
    To make the water colder, energy has to be taken from it, right? So all that thermal energy that was in the water must have been taken from the water and pumped out of the fridge, by the fridge. Would this be the 70% wasted energy?
  6. Apr 25, 2015 #5
    If the fridge was 100% efficient, then would only 1 J be needed? I'm not sure how to answer the question...
  7. Apr 25, 2015 #6


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    I'm not sure how exactly "efficiency" is defined here.
    Do you know the Carnot efficiency? If yes, it is probably the baseline ("100%").

    No, the heat you pump out is the useful part.
  8. Apr 25, 2015 #7
    I'm not familiar with the Carnot efficiency. I'm describing efficiency as per the efficiency equation - Efficiency=EnergyOutput/EnergyInput × 100%. So if the heat that's being pumped out of the fridge is useful, then that's the 30% that is used to cool off the water... This makes sense. But now where does the 70% wasted energy go? Does it escape the system through the walls of the fridge?
  9. Apr 25, 2015 #8


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    Both the energy taken from the electricity grid and the energy taken from the water will heat the environment in some way.
  10. Apr 25, 2015 #9


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    As others have stated, removing 3500 J from this amount of water is way too little energy to produce the temperature (and phase) change desired. You need to show your calculations.

    If this refrigerator is 30% efficient, this means that for every 100 J of energy consumed by the machine, only 30 J of heat are actually removed from inside the frige.
  11. Apr 25, 2015 #10
    I've found what I did wrong - I didn't put a negative sign in front of QF. Now the QTotal equals a much greater negative number, and the equations make sense. Thanks everyone for your help!!
  12. Apr 26, 2015 #11

    Simon Bridge

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    So basically you discovered a procedural flaw in the way you used the equations without understanding the physics behind them?

    Just from the usual definitions of the concepts:
    In a nutshell: the fridge is a machine that uses total energy E to move the heat Q from the inside to the outside.(Note: This statement defines E and Q here.)
    The efficiency (as a percentage) would be e=100xQ/E so, given Q, E is given by: E=100Q/e. You are told the efficiency.
    If the fridge operates with power P, then it takes T=E/P time to move that Q heat.
    Notice - no hard memorizing or getting the right minus signs or subscripts etc needed.

    At 100% efficiency, E=Q so T=Q/P ... in real life this won't be possible. The maximum possible efficiency is less than 100% - usually lots less. But your fridge is not assumed to be maximally efficient - and the efficiency is given - I just wanted to check your concepts. Since then you had other feedback.

    Ideally you should sketch a heat-flow diagram and analyse the fridge as a heat engine. Check: does that make a difference to the description above?
    (Though I suspect you have not covered "heat engines" yet.)

    Still not quite right because the heat Q starts out in the water, not floating about in the fridge compartment. The fridge works to keep the temperature at the location of the thermostat at a sort-of constant Tc, but the rate that heat from the water moves into the fridge compartment depends on the rate of cooling ... which does not depend just on the power of the fridge. If the fridge has some sort of fan-convection circulating the air inside, then the rate of cooling may be reasonably well approximated by Newton's law of cooling... otherwise it will probably be slower. Much depends on the details - but this is what I was talking about when I said that I thought there was something wrong with the question. It's also what happens when you know more physics than the problem is testing for. The upshot is that E/P is likely the quickest the fridge could cool the water to it's operating temperature.

    Note: I make the Carnot (= max possible) efficiency of this machine at about 12% using Th="room temp" as 20deg.
    Unless we take it that the outside temp is about 90degC...

    Anyway: all the problem actually wants you to do is show that you understand about energy, efficiency, and power.
    Just using a bunch of equations you have memorized does not show that you understand the physics but you may get away with it anyway - good luck.
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