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Time for temperature of water to raise

  1. Sep 30, 2014 #1
    1. The problem statement, all variables and given/known data
    How long will it take to heat 40 gallons of water from 70°F to 120°F with a 20 kW immersion heater assuming no energy is lost the environment?

    2. Relevant equations
    It takes an electric tea kettle with 20°C water five minutes to reach boiling at 100°C. How long will it take for all the water to boil away assuming the same rate of heat addition?

    3. The attempt at a solution
    I tried to find BTU to change to kWh. I'm really not sure which equations should be used. My book provides (k x A x (T2-T1))/elapsed time. It also provides Q = mc(change of temp). There's no other equations
     
  2. jcsd
  3. Oct 1, 2014 #2

    rude man

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    First, change all units as necessary to SI as youhave suggested. Look up the conversions on the Web.

    1: Q = CpΔT. Take d/dt of both sides to solve problem 1.

    2: First, assume 1 mole of water & compute Q' = dQ/dt required to bring water from 20C to 100C in 5 minutes.
    Then, calculate heat required to boil all the water away. Hint: heat of vaporization.
    Finally, divide each side of the equation by Δt and solve for Δt.
     
  4. Oct 1, 2014 #3
    I'm extremely confused. The answer that the professor gave me was 14.67 minutes. I keep ending up at 33.9 although I've used the equation in several ways. I converted 40 gallons to 33.8 kg and change of temp to 10 degrees Celsius. I'm not sure what to do now...
     
  5. Oct 1, 2014 #4

    rude man

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    Which problem are you solving? 1 or 2?
    40 gal. water = 151.4 kg, not 33.8 kg
    Change of 50F is not 10C!
     
  6. Oct 1, 2014 #5
    The first equation is 14.67 minutes. The second equation is 33.75 minutes. This from my professor which I still do not understand how to come up with this. Heat of vaporization is 540 calories per gram was given as information for the second question.

    For the first question I have done the following:
    Change of temp is 50 degrees Fahrenheit which converted to Celsius is 10 degrees.
    Q = mc(delta)T = 151.41kg * 0.59kW * 10 degrees Celsius = 893 J (energy required)
    Q/t = (k x A x (T2-T1))/elapsed time
    k is thermal conductivity, A is the surface area the thickness which the heat flows through... I don't understand how I would get the time elapsed or where do I go from here.

    For the second equation I have not been able to figure out how to apply the equations at all. The symbols are confusing to me. I am in a sophomore physics class, I'm not doing advanced thermodynamics or the sort.
     
  7. Oct 2, 2014 #6

    haruspex

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    Not so. How are you doing this conversion?
     
  8. Oct 2, 2014 #7
    Here's How:
      • Take the temperature in Fahrenheit subtract 32.
      • Divide by 1.8.
      • The result is degrees Celsius.
    50-32=18/1.8=10
     
  9. Oct 2, 2014 #8

    haruspex

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    That converts a temperature of 50F to a temperature of 10C. But what you have to convert is a temperature difference of 50F. That is not the same thing. E.g. try converting 0F to degrees C, then take that from 10C. (But there is a more direct way.)
     
  10. Oct 2, 2014 #9
    Ok so one of the kids in my class had this for the first question:
    40 gallons * 8.3 lbs/gallons = 332 lbs
    332lbs * 50 deg F = 16,600 Btu = 4.89 kW
    4.89 kWh/20kW = 0.245 Hours or 14 minutes approximately
     
  11. Oct 2, 2014 #10

    haruspex

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    No doubt that's right, but you won't learn much from being told that. More importantly, do you understand why your conversion to degrees C was wrong?
     
  12. Oct 2, 2014 #11
    No, not really. I think he used it because of the Btu conversion part. I think that requires the American system of lbs and Fahrenheit.
     
  13. Oct 2, 2014 #12

    haruspex

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    Yes, you can solve it entirely using US units, but you should get the same answer if you convert all to international standard units first. If you are going to be dealing with US units again, it will be important that you know how to do the conversions.
    If you have a temperature f0 in degrees Fahrenheit then to convert it to degrees Celsius you do as you did: c0 = (f0-32)*5/9. Likewise, if you have a higher temperature f1: c1 = (f1-32)*5/9. But suppose all that you are interested in is converting the temperature difference f1 - f0. What do these two equations give you for c1 - c0?
     
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