Question

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?

Answer 1

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.

Answer 2

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.