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QPage 47: Q16: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q16 Solution: A ball is dropped on a floor from a height of 2.0 m. After the…

A ball is dropped on a floor from a height of 2.0 m. After the collision it rises up to a height of 1.5 m. Assume that 40% of the mechanical energy lost goes as thermal energy into the ball. Calculate the rise in the temperature of the ball in the collision. Heat capacity of the ball is 800 J K−1800 \text{ J K}^{-1}.

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**Answer:** Rise in temperature = 2.5×10−32.5 \times 10^{-3} �C. **Step-by-Step Explanation:** Given: Height dropped = 2.0 m; Height rebounded = 1.5 m; 40% of mechanical energy lost goes as thermal energy into ball; Heat capacity of ball = 800 J/K. Mechanical energy lost = mg(h1−h2)=m×10×(2.0−1.5)=5mmg(h_1 - h_2) = m \times 10 \times (2.0 - 1.5) = 5m J. Thermal energy into ball = 0.4×5m=2m0.4 \times 5m = 2m J. Rise in temperature: 2m=800×ΔT⇒ΔT=2m/8002m = 800 \times \Delta T \Rightarrow \Delta T = 2m/800. For m=1m = 1 kg (unit mass): ΔT=0.0025�C=2.5×10−3�C\Delta T = 0.0025�\text{C} = 2.5 \times 10^{-3}�\text{C}.[citation:6]

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