Heat capacity of a calorimeter
Karl's comment (see above) prompt me to suggest the following rough experiment which can be done in the classroom with minimum equipment:
Suppose you are required to find out the specific heat capacity of water. Weigh $m_1 \ \pu{g}$ of warm water at $T_1 \ \pu{^\circ C}$ to a Styrofoam cup with a lid. Also weigh $m_2 \ \pu{g}$ of ice at $T_2 \ \pu{^\circ C}$ to another Styrofoam cup. Then add ice to warm water (Note: Measure the temperature of warm water just before addition to make sure it is still at $T_1 \ \pu{^\circ C}$). Stir the mixture carefully while the coup is covered by the lid. Measure the temperature of the mixture until reading is steady. Suppose the steady temperature is $T_3 \ \pu{^\circ C}$. Ensure this time no ice is visible (meaning all dissolved). Now we are ready for the calculations:
Suppose the specific heat capacity of water is $c_{(\ell)}$ and the specific heat capacity of water is $c_{(s)}$. For novices, we can assume that $c_{(\ell)} \approx c_{(s)} = c$. Thus, heat lost by warm water at equilibrium:
$$Q_\text{lost} = m_1 \times c \times (T_1 - T_3) \tag1$$
For ice, if $T_2 \lt 0$, ice would gain heat first to become ice at $\pu{0 ^\circ C}$, then $\pu{0 ^\circ C}$ ice melts to $\pu{0 ^\circ C}$ water, and warm up to $T_3 \ \pu{^\circ C}$ water. If heat of fusion of ice is $L \ \pu{Jg-1}$, total heat gained by ice at equilibrium:
$$Q_\text{gained} = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) \tag2$$
At equilibrium, $Q_\text{gained} =  Q_\text{lost}$. Therefore, equations $(1)$ and $(2)$:
$$m_1 \times c \times (T_1 - T_3) = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) tag3$$
Only unknown in the equation $(3)$ is $c$, you can solve the equation $(3)$ for $c$.
Also note that you can avoid $m_2 \times c \times (0 - T_2)$ part by choosing $\pu{0 ^\circ C}$ ice, because it is easy to get $\pu{-20 ^\circ C}$ from freezer, it it kept in constant temperature.