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Question
An electric room heater has a resistance of 40 Ω and operates at 220 V for 10 min. Calculate the heat energy dissipated by it in kilojoule.
Numerical
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Solution
R = 40 Q; V = 220 V; t = 10 × 60 = 600 s
\[\text{Energy dissipated} = \frac{\mathrm{V}^2 \times t}{\mathrm{R}}\]
\[= \frac{220 \times 220 \times 600}{40}\]
= 726000 J
= 726.0 kJ
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