I have to calculate the Ccal using the following info:

Question

Initial Temp of water outside calorimeter=22.6 C
Initial Temp of water inside the calorimeter= 100 C
Volume of water in calorimeter= 50mL
Volume of water outside calorimeter= 300mL
Final Temp of water in calorimeter=27.9 C
Final Temp of water outside calorimeter= 27.9 C

Answer 1

To calculate the heat capacity of the calorimeter (C_cal), we will first need to calculate the heat gained by the water outside the calorimeter (Q1) and the heat lost by the water inside the calorimeter (Q2). Then we can determine C_cal using C_cal = (Q2 - Q1) / ΔT.

Step 1: Calculate Q1 (heat gained by water outside the calorimeter)
Q1 = m1 * C_w * ΔT1
where m1 is the mass of water outside the calorimeter, C_w is the specific heat capacity of water (4.18 J/g°C), and ΔT1 is the change in temperature of water outside the calorimeter.

m1 = volume * density = 300 mL * 1 g/mL = 300 g
ΔT1 = T_final - T_initial = 27.9°C - 22.6°C = 5.3°C

Q1 = 300 g * 4.18 J/g°C * 5.3°C = 6631.56 J

Step 2: Calculate Q2 (heat lost by water inside the calorimeter)
Q2 = m2 * C_w * ΔT2
where m2 is the mass of water inside the calorimeter, and ΔT2 is the change in temperature of water inside the calorimeter.

m2 = volume * density = 50 mL * 1 g/mL = 50 g
ΔT2 = T_initial - T_final = 100°C - 27.9°C = 72.1°C

Q2 = 50 g * 4.18 J/g°C * 72.1°C = 15082.26 J

Step 3: Calculate C_cal (heat capacity of the calorimeter)
C_cal = (Q2 - Q1) / ΔT_combination
where ΔT_combination is the change in temperature for the combined water.

ΔT_combination = |ΔT1| = 5.3°C (since the final temperatures are equal)

C_cal = (15082.26 J - 6631.56 J) / 5.3°C = 8440.70 J / 5.3°C = 1592.58 J/°C

The heat capacity of the calorimeter (C_cal) is approximately 1592.58 J/°C.