Answer
The initial temperature of the lead ball was $452~^{\circ}C$
Work Step by Step
We can find the mass of the water as:
$m_w = \rho~V$
$m_w = (1000~kg/m^3)(2.5~L)(\frac{1~m^3}{1000~L})$
$m_w = 2.5~kg$
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
The heat energy lost by the lead ball will be equal in magnitude to the heat energy gained by the water. We can find the initial temperature of the lead ball.
$m_l~c_l~\Delta T_l = m_w~c_w~\Delta T_w$
$m_l~c_l~(T-32.0^{\circ}C) = m_w~c_w~\Delta T_w$
$m_l~c_l~T = m_w~c_w~\Delta T_w+m_l~c_l~(32.0^{\circ}C)$
$T = \frac{m_w~c_w~\Delta T_w+m_l~c_l~(32.0^{\circ}C)}{m_l~c_l}$
$T = \frac{(2.5~kg)(4186~J/kg~C^{\circ})(12.0~C^{\circ})+(2.3~kg)(130~J/kg~C^{\circ})~(32.0^{\circ}C)}{(2.3~kg)(130~J/kg~C^{\circ})}$
$T = 452~^{\circ}C$
The initial temperature of the lead ball was $452~^{\circ}C$.
