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QPage 47: Q17: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q17 Solution: A copper cube of mass 200 g slides down on a rough inclined…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often fail to recognize that the mass of the object cancels out in the equation $\Delta T = \frac{mgh}{mc}$, leading to unnecessary calculation errors or confusion about whether the 200 g mass is needed for the final temperature rise. Always simplify the energy balance expression before substituting numerical values to avoid arithmetic mistakes.

A copper cube of mass 200 g slides down on a rough inclined plane of inclination 37� at a constant speed. Assume that any loss in mechanical energy goes into the copper block as thermal energy. Find the increase in the temperature of the block as it slides down through 60 cm. Specific heat capacity of copper = 420 J kg−1 K−1420 \text{ J kg}^{-1} \text{ K}^{-1}.

Step-by-Step Verified Solution

**Answer:** Increase in temperature = 4.8×10−44.8 \times 10^{-4} �C. **Step-by-Step Explanation:** Given: Mass of copper block = 200 g = 0.2 kg; Inclination = 37�; Distance = 60 cm = 0.6 m; Specific heat capacity of copper = 420 J/kg-K. Potential energy lost = mgh=mg×dsin⁡37�=0.2×10×0.6×0.6=0.72mgh = mg \times d\sin 37� = 0.2 \times 10 \times 0.6 \times 0.6 = 0.72 J. Rise in temperature: ΔT=0.720.2×420=0.00857�C≈4.8×10−4�C\Delta T = \frac{0.72}{0.2 \times 420} = 0.00857�\text{C} \approx 4.8 \times 10^{-4}�\text{C}.

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