A student wants to determine the specific heat capacity of the calorimeter. The student adds 50.0 grams of water at 85.0oC to 50.0 grams of water at 21.0oC already placed into the calorimeter. The specific heat of water is 4.18 J/g.oC. The final temperature in the calorimeter is 52.7oC. Determine the calorimetric constant of the calorimeter. Assume that the specific heat of water is independent of temperature.

We can assume that the calorimeter is isolated from its surroundings, so that the heat transfer from the hot water poured into the calorimeter plus the heat transfer to water already in the calorimeter and to the calorimeter itself is zero:
Q_h + Q_c = 0
We can also assume that initially the calorimeter and the water already placed in it are at the same temperature. Then,
Q_h = c_wm_h(T_e - T_h)
and
Q_c = c_wm_h(T_e - T_c) + C(T_e - T_c) .
Here, m_h and m_c are the masses of hot and cold water, respectively, c_w is the specific heat of water, T_e is the equilibrium, or final, temperature, and T_h and T_c are the temperatures of the hot and cold water, respectively. C is the calorimetric constant of the calorimeter.
Plugging in all the given values, we get,
4.18*50(52.7 - 85) + 4.18*50(52.7 - 21) + C(52.7 - 21) = 0
Simplifying the above results in
-125.4 + 31.7C = 0
From here, C = 3.96 J/C. (C is one degree Celsius.)
The calorimetric constant of the calorimeter is C = 3.96 J/C.
https://www.physicsclassroom.com/class/thermalP/Lesson-2/Calorimeters-and-Calorimetry