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Resistance of a water heaters heating element?

  1. Jul 23, 2017 #1
    1. The problem statement, all variables and given/known data
    Household water heaters use a 240 V rather than a 120 V source. What is the resistance of a water heater's heating element if it heats 40 gallons (151 kg) of water from 15 C to 60 C in 15 min

    2. Relevant equations
    P = IV
    P = I^2R
    I = dq/dt
    3. The attempt at a solution
    Honestly no clue. If anyone can give me some hints I would really appreciate this.

    So far I have electrical units, V, and then I have kg, Celcius, and seconds. I'm not sure how the given equations help me relate the units at all.

    I need current and power, but those units do not help me at all.

    I know I = dq/dt, and I thought 45 C / 900 s looks very similar to current but the units don't make sense to me...
     
  2. jcsd
  3. Jul 23, 2017 #2
    Do you know how to determine how much heat it takes to raise the temperature of 1 kg of water from 15 C to 60 C?
     
  4. Jul 23, 2017 #3
    I'm not sure actually.. this chapter that I'm studying is called Current and Resistance..

    It has to do with specific heat, right? Also, does heat = thermal energy?
     
  5. Jul 23, 2017 #4
    Yes. Did you not study heat in freshman physics?
     
  6. Jul 23, 2017 #5
    I don't think I studied it enough..

    So if the specific heat of water is 4.187 x 10^3 J/(kg * k )

    That means it requires 4.187 x 10^3 J to raise 1 kg of water by 1 kelvin, correct?
     
  7. Jul 23, 2017 #6

    scottdave

    User Avatar
    Homework Helper
    Gold Member

    Yes, heat = thermal energy. To raise a kg water 1°C, you need 4184 Joules of heat (thermal energy).

    http://www.kentchemistry.com/links/Energy/SpecificHeat.htm
     
  8. Jul 23, 2017 #7
    What is the mass in kg of 40 gallons of water?
     
    Last edited: Jul 23, 2017
  9. Jul 23, 2017 #8
    I think the asker had the converted number already . They wrote 151 kg.
     
  10. Jul 23, 2017 #9
    Anyhow, using the heat you can now calculate the power (J/s). See that the heat is the 'J' and the time you have the 's'.

    Now think about how you can get current!
     
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