Answer
The initial temperature of the horseshoe is $174~^{\circ}C$
Work Step by Step
We can find the mass of the water as:
$m = \rho~V$
$m = (1000~kg/m^3)(1.25~L)(\frac{10^{-3}~m^3}{1~L})$
$m = 1.25~kg$
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
We then find the heat energy gained by the water and the iron pot:
$Q = m_w~c_w~\Delta T + m_i~c_i~\Delta T$
$Q = (1.25~kg)(4186~J/kg~C^{\circ})(5.0~C^{\circ}) + (0.30~kg)(450~J/kg~C^{\circ})(5.0~C^{\circ})$
$Q = 26,837.5~J$
The heat energy lost by the iron horseshoe will be equal in magnitude to the heat energy gained by the water and the iron pot. We can find the initial temperature $T_i$ of the horseshoe.
$Q = m_i~c_i~(T_i-25^{\circ}C)$
$m_i~c_i~T_i = Q + m_i~c_i ~(25^{\circ}C)$
$T_i = \frac{Q + m_i~c_i ~(25^{\circ}C)}{m_i~c_i}$
$T_i = \frac{26,837.5~J + (0.40~kg)(450~J/kg~C^{\circ})(25^{\circ}C)}{(0.40~kg)(450~J/kg~C^{\circ})}$
$T_i = 174~^{\circ}C$
The initial temperature of the horseshoe is $174~^{\circ}C$.
