NEET-XII-Physics
25: Calorimetry
- #16A 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-1.Ans : Height of the floor from which ball is dropped, h1 = 2.0 m
Height to which the ball rises after collision, h2 = 1.5 m
Let the mass of ball be m kg.
Let the speed of the ball when it falls from h1 and h2 be v1 and v2, respectively.
`` {v}_{1}=\sqrt{2g{h}_{1}}=\sqrt{2\times 10\times 2}=\sqrt{40}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
`` {v}_{2}=\sqrt{2g{h}_{2}}=\sqrt{2\times 10\times 1.5}=\sqrt{30}\,\mathrm{\,m\,}/\,\mathrm{\,s\,}``
Change in kinetic energy is given by
`` ∆K=\frac{1}{2}\times \,\mathrm{\,m\,}\times 40-\left(\frac{1}{2}\,\mathrm{\,m\,}\right)\times 30=\left(\frac{10}{2}\right)\,\mathrm{\,m\,}``
`` \Rightarrow ∆K=5\,\mathrm{\,m\,}``
If the position of the ball is considered just before hitting the ground and after its first collision, then 40% of the change in its KE will give the change in thermal energy of the ball. At these positions, the PE of the ball is same. Thus,
Loss in PE = 0
The change in kinetic energy is utilised in increasing the temperature of the ball.
Let the change in temperature be ΔT. Then,
`` \left(\frac{40}{100}\right)\times ∆K=m\times 800\times ∆T``
`` \left(\frac{40}{100}\right)\times \frac{10}{2}m=m\times 800\times ∆T``
`` \Rightarrow ∆T=\frac{1}{400}=0.0025``
`` =2.5\times {10}^{-3}°\,\mathrm{\,C\,}``
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