#### Answer

The initial temperature of the pan was $272^{\circ}C$ which is $522^{\circ}F$

#### Work Step by Step

We can find the mass of the water as:
$m_w = \rho~V$
$m_w = (1000~kg/m^3)(10.0~L)(\frac{1~m^3}{10^3~L})$
$m_w = 10.0~kg$
The heat energy lost by the aluminum will be equal to the heat energy gained by the water. We can find the initial temperature of the pan:
$m_a~c_a~\Delta T_a = m_w~c_w~\Delta T_w$
$m_a~c_a~(T - 24.0^{\circ}C) = m_w~c_w~\Delta T_w$
$m_a~c_a~T = m_w~c_w~\Delta T_w+m_a~c_a~(24.0^{\circ}C)$
$T = \frac{m_w~c_w~\Delta T_w+m_a~c_a~(24.0^{\circ}C)}{m_a~c_a}$
$T = \frac{(10.0~kg)(4186~J/kg~C^{\circ})(4.0~C^{\circ})+(0.750~kg)(900~J/kg~C^{\circ})(24.0^{\circ}C)}{(0.750~kg)(900~J/kg~C^{\circ})}$
$T = 272~^{\circ}C$
We can convert this temperature to Fahrenheit.
$F = \frac{9}{5}C+32$
$F = \frac{9}{5}(272)+32$
$F = 522$
The initial temperature of the pan was $272^{\circ}C$ which is $522^{\circ}F$.