what is the required resistance of an immersion heater that will increase the temperature of 1.50 kg of water from 10 degrees C to 50 degrees C in 10 minutes while operating 120 V?

Answer 1

heat required=1.5kg*cwater*(50-10) joules

power required= heat/time=heat/600sec

v^2/r= power

resistance=120^2*600/(1.5kj(cwater)40) ohms

check that.