A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is
(Specific heat of steel $= 460 Joule kg^{-1}°C^{-1}, g = 10 ms^{-2}$ )
Text Solution
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According to energy conservation, change in potential energy of the ball, appears in the form
of heat which raises the temperature of the ball.
i.e. $mg(h_1 - h_2) = m.c.\Delta \theta$
⇒ ⇒ $\Delta \theta = \frac{g(h_1 - h_2)}{c}$
$\frac{10(10-5.4)}{460} = 0.1^\circ \mathrm{C}$
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