Posted by Machelle Butts on .
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,
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,
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.