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QPage 47: Q8: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q8 Solution: Calculate the time required to heat 20 kg of water from 10�C to …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently calculate time using the full rated power ($1000 \text{ W}$) instead of the effective power, ignoring the $80\%$ efficiency. Always multiply the rated power by the efficiency factor to find the actual power available for heating before dividing the total heat energy by it.

Calculate the time required to heat 20 kg of water from 10�C to 35�C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1}.

Step-by-Step Verified Solution

**Answer:** The time required is 2625 seconds � 43.75 minutes. **Step-by-Step Explanation:** Given: Mass of water = 20 kg; Initial temp = 10�C; Final temp = 35�C; Power of heater = 1000 W; Efficiency = 80%; Specific heat capacity of water = 4200 J/kg-K. Heat required:
Q=20×4200×(35−10)=20×4200×25=2.1×106 JQ = 20 \times 4200 \times (35 - 10) = 20 \times 4200 \times 25 = 2.1 \times 10^6 \text{ J}
Useful power = 0.8×1000=8000.8 \times 1000 = 800 W. Time required:
t=2.1×106800=2625 st = \frac{2.1 \times 10^6}{800} = 2625 \text{ s}

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