Posted by **Anonymous** on Saturday, February 23, 2008 at 3:29pm.

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 -
**Damon**, Saturday, February 23, 2008 at 5:02pm
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

so

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

So

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

- physics -
**drwls**, Saturday, February 23, 2008 at 5:11pm
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.

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