Linking Electricity and Thermodynamics
Level
pending
An immersion heater must increase the temperature of 1.50 kg of water from 10.0°C to 50.0°C in 10.0 min while operating at 110 V. What is the required resistance of the heater? hint: rate of energy Q= Mc*(delta)T
The answer is 29.
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1 solution
Q=M* C * dT => 1.5 * 4200 * (50-10) = 252000 J
now ,
252000J = V * I * t = 110 * I * 10x60 sec. ==> I= 3.8181 Amps.
R= V/I
R= 110/3.8181 = 28.8 Ohms ~ 29 Ohms.