Post a New Question


posted by .

What is the required resistance of an immersion heater that will increase the temperature of 1.5kg of water from 10*C to 50*C in 10 min while opperating at 120V? The answer is 24.2 ohms. I don't know what equation to use to solve this question. Please help me! Thanks!

  • physics -

    Power = V^2/R = 120^2/R = 1.44*10^4/R
    Energy = power * time
    where time = 10 minutes = 600 seconds = 6*10^2 s
    Energy in Joules = (1.44*6)*10^6/R
    = 8.64*10^6/R in Joules

    Now, how many Joules does it take to heat the water?
    Joules into water = 1.5 kg*4190 J/kg deg*(50-10) deg = 251,400 Joules = 2.51*10^5 Joules

    8.64 *10^6 / R = 2.51 * 10^5

  • physics -

    Don't try to find one magical formula that you just plug into to get the answer. Do the problem one step at a time and use formulas that you understand, your powers of reasoning. That is how physics and engineering is done.

    The heat energy that must be added to the water is
    Q = M C *(delta T)
    = 1500 g*4.184 J/(g C)*40 C = 2.51*10^5 J
    delta T is the temperature change and C is the specific heat of water.

    To provide this much heat in T = 10 min (which is 600 s), the electrical power must be
    P = Q/T = 2.51*10^5/600 = 415 J/s or watts

    The last step is choosing the right resistor R, is to use
    P = V^2/R, where V is the voltage and P is the required power. The answer will be in ohms.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question