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What is the resistance of the heater?

  1. Jun 5, 2006 #1
    "A student uses an immersion heater to heat a cup of water (300mL) from 20ºC to 80ºC for tea. If it is 75% efficient and takes 2.5 min, what is the resistance of the heater? (Assume 120-V household voltage.)"

    Here's what I did:
    V= 0.75 * 120 = 90V
    mass of water = 0.3 kg, specific heat c = 4186 J/(kg*Cº)
    T = change in temp = 60ºC
    t = 2.5 min = 150s

    Q (heat) = W (work) = mcT = 75348 J
    P = W/t = 502.32 W
    P = 0.75 * P = 376.74 W
    P = V2/R
    R = V2/P = 21.5 ohms

    First, are my values for voltage and power correct? I wasn't sure where to apply the 75% efficiency.

    Second, is my answer correct? it's supposed to be 21 ohms... I changed some values to have exactly 2 significant figures, but my answer just gets farther away from 21. (for example, I tried changing 75348 J to 75000J, but that makes the answer 21.6 ohms, which is farther away from 21 ohms...)
     
  2. jcsd
  3. Jun 5, 2006 #2

    Andrew Mason

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    No. Since [itex]P = V^2/R \text{ or } P = I^2R[/itex], the energy consumed by the resistance is proportional to the square of the voltage or current. If it is 75% efficient, then [itex]E = .75 * I^2R\Delta t[/itex]

    Also, who drinks tea made with 80ºC water? You need to bring it to a boil!

    AM
     
    Last edited: Jun 5, 2006
  4. Jun 5, 2006 #3
    So [tex]R = .75 * \frac{V^2}{E}\Delta t[/tex]
    but Power = E / time, so
    [tex]R = .75 * \frac{V^2}{P}[/tex]
    I arrive at the same answer (21.5 ohms) however, but that's probably a coincidence?
     
  5. Jun 5, 2006 #4

    Andrew Mason

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    Your answer is right actually because you didn't use V=90 volts in your calculation.

    AM
     
  6. Jul 12, 2006 #5
    I also got R=21.5 ohms. First, get the total energy needed to boil the water from 20deg to 80deg. Second, take note that energy per 2.5min is only 75% so u need to make it 100%. Finally, apply R =V^2/P to get R.
     
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