posted by Kunal on .
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?
First calculate the heat energy Q needed fr that amount of heating
Q = M C *(delta T) = 1500g*1.00 cal/gm C*40 C = 6*10^4 cal = 2.51*10^5 J
Then require that the electrical power be such that
(V^2/R)*600 s = 2.51*10^5 J
and solve for R