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
Correct Answer: Option D
Explanation
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