Answer
The original temperature of the water was $42.7~^{\circ}C$
Work Step by Step
We can find the mass of the water as:
$m = \rho~V$
$m = (1000~kg/m^3)(135~mL)(\frac{10^{-6}~m^3}{1~mL})$
$m = 0.135~kg$
The energy $Q$ required to raise the temperature of a substance is:
$Q = mc~\Delta T$
Let $T_e$ be the equilibrium temperature. The heat energy lost by the water will be equal in magnitude to the heat energy gained by the glass. We can find the original temperature $T_w$ of the water.
$m_w~c_w~(T_w-T_e) = m_g~c_g~(T_e-T_g)$
$m_w~c_w~T_w = m_g~c_g~(T_e-T_g)+m_w~c_w~T_e$
$T_w = \frac{m_g~c_g~(T_e-T_g)+m_w~c_w~T_e}{m_w~c_w}$
$T_w = \frac{(0.0315~kg)(840~J/kg~C^{\circ})~(41.8^{\circ}C-23.6^{\circ}C)+(0.135~kg)(4186~J/kg~C^{\circ})(41.8^{\circ}C)}{(0.135~kg)(4186~J/kg~C^{\circ})}$
$T_w = 42.7~^{\circ}C$
The original temperature of the water was $42.7~^{\circ}C$.
