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10-1.Thermometry, Thermal Expansion and Calorimetry
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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 ........... $^\circ \mathrm{C}$ (Specific heat of steel $ = 460\,Joule - k{g^{ - 1}}^\circ {C^{ - 1}},\;g = 10\,m{s^{ - 2}}$)
A
$0.01$
B
$0.1$
C
$1$
D
$1.1$
Solution
(b) 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 $
==> $\Delta \theta = \frac{{g({h_1} – {h_2})}}{S}$
$ = \frac{{10(10 – 5.4)}}{{460}} = 0.1^\circ C$
Standard 11
Physics
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