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QPages 218-219: Q9: HC Verma: Concepts of Physics - Volume 2 Thermal and Chemical Effects of Electric Current Pages 218-219: Q9 Solution: An immersion heater rated 1000 W, 220 V is used to heat of…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently apply the 60% efficiency to the voltage or resistance (e.g., using 0.6 V or 0.6 R) instead of multiplying the total electrical power by 0.6, which double‑counts the loss and yields a wrong useful power. Compute the heater’s electrical power (1000 W or V²/R) first, then apply the 60% efficiency to obtain the actual heating power.

An immersion heater rated 1000 W, 220 V is used to heat 0.01 m30.01 \text{ m}^3 of water. Assuming that the power is supplied at 220 V and 60% of the power supplied is used to heat the water, how long will it take to increase the temperature of the water from 15�C to 40�C?

Step-by-Step Verified Solution

**Answer:** t=29.16t = 29.16 min. **Step-by-Step Explanation:** R=2202/1000=48.4 ΩR = 220^2/1000 = 48.4\ \Omega. Mass of water = 10 kg. Heat needed: Q=10×4200×25=1.05×106Q = 10 \times 4200 \times 25 = 1.05 \times 10^6 J. Useful power = 0.6×(2202/48.4)=6000.6 \times (220^2/48.4) = 600 W. t=1.05×106/600=1750t = 1.05 \times 10^6 / 600 = 1750 s = 29.16 min .

All Questions in this Chapter

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