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What is the required resistance of an immersion heater that will increase the temperature of 2.00 kg of water from 10.0°C to 50.0°C in 9.0 min while operating at 120 V?

  • physics -

    Electrical energy in
    = (V^2/R)*time (in Joules)
    (use time = 540 seconds)
    All of this energy is converted to heat, Q.

    The electrical energy required is
    Q = M C *(delta T)
    = 2000 g*[4.186 J/(deg C*g)]*40 C

    Use those two equations to solve for R.

  • physics -

    Electrical energy in
    = (V^2/R)*time (in Joules)
    (use time = 540 seconds)
    All of this energy is converted to heat, Q.

    The electrical energy required is
    Q = M C *(delta T)
    = 2000 g*[4.186 J/(deg C*g)]*40 C

    Use those two equations to solve for R.

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