JAMB - Physics (1995 - No. 34)
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
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
heat energy= I2RT
= 2 X 2 X 50 X 200
= 40000J
to convert to kJ, divide J by 1000
hence, \(\frac{40000}{1000}\) = 40kJ
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