Answer
The sphere's temperature will rise $0.39~C^{\circ}$
Work Step by Step
The amount of thermal energy produced when the sphere hits the ground will be equal to the sphere's kinetic energy, which will be equal to the initial potential energy. We can find the potential energy of the sphere as:
$PE = mgh$
$PE = (0.095~kg)(9.80~m/s^2)(55~m)$
$PE = 51.2~J$
We can use this equation to find the temperature increase of the sphere:
$Q = mc~\Delta T$
In this case, $Q$ will be equal to 65% of the potential energy. We can find $Q$:
$Q = 0.65~PE$
$Q = (0.65)(51.2~J)$
$Q = 33.3~J$
We can find the temperature rise of the sphere as:
$Q = m_a~c_a~\Delta T$
$\Delta T = \frac{Q}{m_a~c_a}$
$\Delta T = \frac{33.3~J}{(0.095~kg)(900~J/kg~C^{\circ})}$
$\Delta T = 0.39~C^{\circ}$
The sphere's temperature will rise $0.39~C^{\circ}$.
