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QPage 47: Q12: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q12 Solution: A brick weighing 4.0 kg is dropped into a 1.0 m deep river from …

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently calculate potential energy using only the drop height ($2.0$ m) and ignore the river depth ($1.0$ m), failing to recognize that the brick continues to descend through the water. The total vertical displacement for the potential energy change is the sum of the air drop and the water depth ($h_{total} = 2.0 + 1.0 = 3.0$ m).

A brick weighing 4.0 kg is dropped into a 1.0 m deep river from a height of 2.0 m. Assuming that 80% of the gravitational potential energy is finally converted into thermal energy, find this thermal energy in calorie.

Step-by-Step Verified Solution

**Answer:** Thermal energy in calories = 23 cal. **Step-by-Step Explanation:** Given: Mass of brick = 4 kg; Total height = 2 + 1 = 3 m;

g = 10

m/s_; 80% of potential energy converted to thermal energy. Total change in potential energy:

PE = mgh = 4 \times 10 \times 3 = 120 \text{ J}

Thermal energy =

0.8 \times 120 = 96

J. In calories:

96/4.186 = 22.9 \approx 23

cal.[citation:2]

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