Physics
posted by Mini .
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?

heat required=1.5kg*cwater*(5010) joules
power required= heat/time=heat/600sec
v^2/r= power
resistance=120^2*600/(1.5kj(cwater)40) ohms
check that.
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