Question

Determine the heat energy generated in terms of kJ in an induction stove when a current of 8 A passes through its coils for 5 min when applied voltage is 230 V.

• A. 525 kJ
• B. 55200 kJ
• C. 552 kJ
• D. 55.2 kJ

Answer 1

The correct option is C

552 kJ

Given: current I = 8 A,
time t = 5 min = 5 $\times$ 60 = 300 s,
voltage V = 230V

We know that, heat generated H = VIt

So, H = 230 $\times$ 8 $\times$ 300

i.e. H = 552000 J

Since, 1 kJ = 1000 J, then 552000 J = $\frac{552000}{1000}$ = 552 kJ