#### Answer

The new temperature of the water is $27.6~^{\circ}C$

#### Work Step by Step

We can find the mass of the water as:
$m_w = \rho~V$
$m_w = (1000~kg/m^3)(100~mL)(\frac{1~m^3}{10^6~mL})$
$m_w = 0.10~kg$
The heat energy lost by the copper pellets will be equal to the heat energy gained by the water. We can find the equilibrium temperature $T_e$:
$m_c~c_c~\Delta T_c = m_w~c_w~\Delta T_w$
$m_c~c_c~(300^{\circ}C-T_e) = m_w~c_w~(T_e-20^{\circ}C)$
$(m_c~c_c+ m_w~c_w)~T_e = m_c~c_c~(300^{\circ}C)+ m_w~c_w~(20^{\circ}C)$
$T_e = \frac{m_c~c_c~(300^{\circ}C)+ m_w~c_w~(20^{\circ}C)}{(m_c~c_c+ m_w~c_w)}$
$T_e = \frac{(0.030~kg)(390~J/kg~C^{\circ})(300^{\circ}C)+ (0.10~kg)(4186~J/kg~C^{\circ})(20^{\circ}C)}{(0.030~kg)(390~J/kg~C^{\circ})+ (0.10~kg)(4186~J/kg~C^{\circ})}$
$T_e = 27.6~^{\circ}C$
The new temperature of the water is $27.6~^{\circ}C$.