Answer
The specific heat of the glycerin is $2300~J/kg~C^{\circ}$
Work Step by Step
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
We can find the magnitude of the heat energy lost by the piece of iron:
$Q = m_i~c_i~\Delta T$
$Q = (0.290~kg)(450~J/kg~C^{\circ})(180^{\circ}C - 38^{\circ}C)$
$Q = 18,531~J$
The heat energy lost by the iron will be equal in magnitude to the heat energy gained by the aluminum and the glycerin. We can find the specific heat $c_g$ of the glycerin.
$Q = m_a~c_a~\Delta T + m_g~c_g~\Delta T$
$m_g~c_g~\Delta T = Q - m_a~c_a~\Delta T$
$c_g = \frac{Q - m_a~c_a~\Delta T}{m_g~\Delta T}$
$c_g = \frac{18,531~J - (0.095~kg)(900~J/kg~C^{\circ})(38^{\circ}C-10^{\circ}C)}{(0.250~kg)(38^{\circ}C-10^{\circ}C)}$
$c_g = 2300~J/kg~C^{\circ}$
The specific heat of the glycerin is $2300~J/kg~C^{\circ}$.
