Calculate Part E all the information is given
For the first trial
calculate the heat gained by the cold water
Q cold = mass * specific heat * (Temperature difference)
Q cold = 50.1927 * 4.184 * (48.9 -19.7) =6132.18 Joules
calculate the heat lost by the hot water
mass of hot water = mass of total water - mass of cold water
mass of hot water = 75.22 - 50.197 = 25.0273 C
Q hot = 25.0273 * 4.184 * (48.9 - 83.5) = -3623.112 Joules
Now apply
Q calorimeter + Q hot + Q cool = 0
Q calorimeter = 3623.112 - 6132.18 = -2509.068 Joules
Q calorimeter = heat capacity * temperature difference
Temperature differece = (48.9 - 19.7) = 29.2 C
Q calorimeter = -2509.068 = heat capacity * 29.2
heat capacity = 2509.068 / 29.2 = 85.92 Joule / gram
Here is an image with the other calculations
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A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 19.09°C and of the hot water, 32.17°C. The maximum temperature of the mixture was found to be 24.55°C. Assume the density of water is 1
1.A student determined the calorimeter constant of the calorimeter, using the procedure described in this module. The student added 50.00 mL of cold water to 50.00 mL of heated, distilled water in a Styrofoam cup. The initial temperature of the cold water was 19.09°C and of the hot water, 32.17°C. The maximum temperature of the mixture was found to be 24.55°C. Assume the density of water is 1