Posted by **Machelle Butts** on Tuesday, February 16, 2010 at 3:40pm.

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 -
**drwls**, Tuesday, February 16, 2010 at 4:03pm
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 -
**sarah**, Friday, January 22, 2016 at 3:27am
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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