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Homework Help: Specific heat

  1. Aug 16, 2016 #1
    1. The problem statement, all variables and given/known data
    If a 5 amp unit can cool 4 cups of water from 75 to 50 degrees fahrenheit in one hour, what temperature could a 30 amp unit cool 128 cups of water in the same amount of time? Is there enough information here to get an idea about what the 30 amp unit would do? If so where would one start?
    2. Relevant equations

    3. The attempt at a solution
    I am not too sure where to start.
  2. jcsd
  3. Aug 16, 2016 #2
    cups are not recognised units....this could be a problem but not necessarily. also degrees fahrenheit is not good but not disasterous
  4. Aug 16, 2016 #3
    so what units could I convert them to so the problem can make more sense?
  5. Aug 16, 2016 #4
    For this problem the unit of volume is immaterial. How much more energy is necessary to cool 128 cups from 75 deg to 50 deg in one hour? How much more power does 30 amps provide compared to 5 amps?
  6. Aug 16, 2016 #5
    So how would I find out how much more energy is necessary to cool 128 cups? An extra 100 watts (The 30 amp unit has an output cooling power of 253 watts compared to the 5 amps which has 154 watts.) So sense the difference is 100 watts, what equation would I use to find out what kind of a difference that would make?
  7. Aug 16, 2016 #6
    From where did you get your power figures? The currents given do not reflect your power difference.
  8. Aug 16, 2016 #7
  9. Aug 16, 2016 #8
    Not to drag this out but where did the currents 5 and 30 amps come from?
  10. Aug 16, 2016 #9
    same box.
  11. Aug 16, 2016 #10


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    Strictly speaking, no. The work needed to cool a body over a given temperature range depends on the ambient temperature.
    If we overlook that, this becomes a very simple question of ratios. With 32 times as much water to cool, how much cooling do you think the 5 amp unit could achieve in the hour?
  12. Aug 16, 2016 #11
    Ok I think I get it. These are two different types of cooler and must be running at different voltages.

    The current spec is immaterial. It is the power spec that is important. To cool anything one must provide energy to do the work of removing heat. If the 5 amps is associated with the 154 watt cooler and the 30 amps with the 253 watt cooler then assuming that they are equally efficient the higher power cooler will remove heat how much faster than the lower power unit? How much more water are you cooling?
  13. Aug 16, 2016 #12
    I would estimate maybe .5 to 1 degrees per hour.
  14. Aug 16, 2016 #13
    I am trying to cool 8 gallons of water, and the speed at which that happens is not too important to me, I was just using 1 hour to try and get some ratio of what I might be able to expect
  15. Aug 16, 2016 #14
    Keep in mind that the currents given in this problem are for two different types of cooling units and not related to actual cooling ability. Use must compare the actual power ratings of the units not the currents. So you use 32 times more energy but do it at a rate that is only about 1.64 ? times the lower power unit.
  16. Aug 16, 2016 #15
    well this is turning out to be a lot more complex than I thought, you lost me on your last sentence.
  17. Aug 16, 2016 #16
    sorry, 8 gals is 32 time greater than 4 cups. so you need to remove 32 time more heat(energy). What takes 1 hr to cool 4 cups will take 32 time longer with the same unit. But the bigger unit has more power and can cool things 253/154 =1.64 times faster.or take only 1/1.64 = 0.61 times as long to cool the 8 gals.
  18. Aug 16, 2016 #17
    so if we make some big assumptions we could guess that if I gave the 8 gals 2 hours to cool with the bigger unit we might be in the 50 degree range?
  19. Aug 16, 2016 #18
    not quite, as an analogy if you took 1 hour to go 4 miles at 4 mph how long would it take for you to go 128 miles at 6.56 (=1.64x4) mph?
  20. Aug 16, 2016 #19
    19.5 hours to travel 128 miles. So it could take all day?
  21. Aug 16, 2016 #20
    That's what I'm thinking. And since cooling device is a plate taking heat only from the bottom you mus take into account that the heat at the top must be transported to the bottom further delaying its removal.
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