An electric current of 2 amperes flows in a heating coil of resistance 50 ohms for 3 minutes 20 seconds. Determine the heat produced
- 0.5kJ
- 8.0kJ
- 20.0kJ
-
40.0kJ
The correct answer is: D
Explanation
Heat energy = Elect. energyheat energy= I2RT
= 2 X 2 X 50 X 200
= 40000J
to convert to kJ, divide J by 1000
hence, \(\frac{40000}{1000}\) = 40kJ