Question

A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, what is the rise in its temperature?
(Specific heat of steel = $460J{{\,}^{{}^\circ }}{{C}^{-1}}k{{g}^{-1}}$ and $g=10m{{s}^{-2}}$)
A). ${{0.01}^{{}^\circ }}C$
B). ${{0.1}^{{}^\circ }}C$
C). ${{1}^{{}^\circ }}C$
D). ${{1.1}^{{}^\circ }}C$

Answer 1

First, find the change in the total energy of the steel ball. Then use the law of conservation of energy to transform potential energy into the internal energy of the ball. Then, use an expression for the change in internal energy to determine the rise in temperature.
Formula used: Gravitational potential energy at height h, $E=mgh$ and change in internal energy, $\Delta U=ms\Delta T$

Complete step-by-step solution:
Gravitational potential energy of the ball at height $h$is given by
$E=mgh$
Where m is the mass of the ball and g is the acceleration due to gravity.
Mass of the ball, $m\text{ }=\text{ }0.1\text{ }kg$, initial height, ${{h}_{i}}=10m$, final height, ${{h}_{f}}=5.4m$
Initially, when the ball is at a height of 10 m, its potential energy is

${{E}_{i}}=mg{{h}_{i}}$
${{E}_{i}}=0.1\times 10\times 10=10J$
Final potential energy when the ball is at height 5.4 m is
${{E}_{f}}=0.1\times 10\times 5.4=5.4J$
Change in potential energy of the ball is
$\Delta E={{E}_{f}}-{{E}_{i}}$
$\Delta E=5.4-10=-4.6J$
The change in potential energy is negative which indicates potential energy decreases.
From the law of conservation of energy, we know that energy can neither be created nor destroyed. Therefore, dissipated energy also changes to some other form.
It is given that all dissipated energy is absorbed by the ball from which we can conclude that the potential energy is changed to the internal energy of the ball. Therefore,
$-\Delta E=\Delta U$
Relation of change in internal energy of a body in terms of its mass (m), specific heat capacity (s), and change in temperature$\Delta T$ is given by
$\Delta U=ms\Delta T$
For the ball, we have $\Delta U=4.6J$ and $s=460Jk{{g}^{-1}}^{{}^\circ }C$
By substituting the data and rearranging, we get
$\Delta T=\dfrac{4.6}{0.1\times 460}$
On solving which, we have
$\Delta T={{0.1}^{{}^\circ }}C$
Hence option B is correct.

Note: The law of conservation of energy states that the total energy of an isolated system remains conserved. This law is valid for all forms of energy.
Change in temperature, whether in Kelvin or degree Celsius is the same. For example, in this problem we have obtained $\Delta T={{0.1}^{{}^\circ }}C$ which is also equal to $0.1K$.