100.0 grams of water at 90.0 celsius is added to 100.0 grams of water at 20.0 celsius in a well insulated calorimeter. What is the final temperature of the water in the calorimeter?

Since the calorimeter is well-insulated, find the heat transfer from the hot water plus the heat transfer to the cold water. As we don't know anything about the calorimeter, we can assume that the heat transferred to the calorimeter in the process is negligible:
Q_h + Q_c = 0
The heat transfer from the hot water is
Q_h = c_wm_h(T_e - T_h) and
the heat transfer to the cold water is
Q_c = c_wm_c(T_e - T_c) .
Here, c_w is the specific heat of water, m_h and m_c are the masses of hot and cold water, respectively, T_h and T_c are the initial temperatures of the hot and cold water, respectively. T_e is the final, or equilibrium, temperature.
Notice that the equilibrium temperature is less than the initial temperature of hot water, making the Q_h negative, and it is greater than the initial temperature of cold water, making Q_c positive.
Plugging in all given values, we get
c_w*100(T_e-90) + c_w*100*(T_e-20) = 0
The specific heat of water and 100 will cancel, so
2T_e - 110 = 0
T_e = 110/2 = 55
The final temperature is 55 degrees Celsius. Notice that because equal amounts of the same substance are being mixed, the final temperature is the average of the initial hot and initial cold temperatures.
https://www.physicsclassroom.com/class/thermalP/Lesson-2/Calorimeters-and-Calorimetry