An electric current of 2 amperes flows in a heating coil of resistance 50 ohms for 3 minutes 20 seconds. Determine the heat produced
A
0.5kJ
B
8.0kJ
C
20.0kJ
D
40.0kJ
Verified Correct Answer: D.
Physical Principles & Numerical Solution
Heat energy = Elect. energy
heat energy= I2RT
= 2 X 2 X 50 X 200
= 40000J
to convert to kJ, divide J by 1000
hence, \(\frac{40000}{1000}\) = 40kJ
Source: Myschool Past Questions Archive (Record #27853)