1. When 50.0 mL of water at 80.0°C was mixed with 50.0 mL of water in a calorimeter at

20.0 C, the final temperature of the 100 mL of water and calorimeter was 47.0 C.

Calculate C cal

The heat capacity of the calorimeter is 5.11 J/g°C

Explanation:

Step 1: Given data

50.0 mL of water with temperature of 80.0 °C

Specific heat capacity of water = 4.184 J/g°C

Consider the density of water = 1g/mL

50.0 mL of water in a calorimeter at 20.0 °C

Final temperature = 47.0 °C

Step 2: Calculate specific heat capacity of the water in calorimeter

Q = Q (cal) + Q (water)

Q (cal) = mass * C (cal) * ΔT

Qwater = mass * Cwater * ΔT

Qcal = - Qwater

mass (cal) * C (cal) * ΔT (cal) = mass (water) * C (water) * ΔT (water)

50 grams * C (cal) * (47.0 - 20.0) = - 50grams * 4.184 J/g°C * (47-80)

1350 * C (cal) = 6903.6

C (cal) = 5.11 J/g°C

The heat capacity of the calorimeter is 5.11 J/g°C