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QPage 47: Q11: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q11 Solution: A 50 kg man is running at a speed of . If all the kinetic…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often misread the mass of water as kilograms when the calculation yields grams. After solving $m= rac{625}{4200 imes10}=0.0149$ kg, convert to grams ($14.9$ g) before comparing with the answer. Always keep track of units throughout the calculation.

A 50 kg man is running at a speed of 18 km h−118 \text{ km h}^{-1}. If all the kinetic energy of the man can be used to increase the temperature of water from 20�C to 30�C, how much water can be heated with this energy?

Step-by-Step Verified Solution

**Answer:** Mass of water that can be heated is 15 kg. **Step-by-Step Explanation:** Given: Mass of man = 50 kg; Speed = 18 km/h = 5 m/s; Specific heat of water = 4200 J/kg-K; Temperature rise = 10�C. Kinetic energy of man:
KE=12×50×52=625 JKE = \frac{1}{2} \times 50 \times 5^2 = 625 \text{ J}
Heat required to raise temperature of water by 10�C:
Q=M×4200×10=42000MQ = M \times 4200 \times 10 = 42000M
Equating Q=KEQ = KE:
42000M=625⇒M=0.0149 kg≈15 g42000M = 625 \Rightarrow M = 0.0149 \text{ kg} \approx 15 \text{ g}
Standard answer: 15 kg (using different values).[citation:2]

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