Question

The Calorimeter Constant is either 22.1J/C or perhaps the specific heat of water can be used? I calculated the answers using both options and neither got me an answer close to the theoretical. This is the given data so I'm not sure why I can't get an answer similar.

PART B Data Analysis: 1. Calculate the heat of dissolving (AH.is) for each trial 2. Calculate the average, standard deviation, and relative percent standard deviation for your three trials 3. Calculate the percent error based on your average and the theoretical value given in your data set. Data Trial 3 45.120 g 105.687 g | Trial 1 Trial 2 Part A (calorimeter constant = 22.1 J/°C) Metal name lead Mass of calorimeter 44.193 g 47.272 g Mass of calorimeter + water 115.543 g | 125.569 g Mass of lead 98.059 g Initial temperature lead 100.0 °C 100.0 °C Initial temperature of water 21.0°C 21.6°C Final temperature 24.0°C 24.2°C Part B Mass of calorimeter | 41.304 g 50.267 g Mass of calorimeter + water 65.390 g 73.507 g mass of cesium hydroxide 3.120 g 5.199 g Initial temperature water/calorimeter 24.8°C 22.0°C Final temperature 36.4 °C 42.2 °C 100.0 °C 23.0°C 26.0°C Trial 1 Trial 2 Trial 3 Part C Volume NaOH 21.2 mL 23.0 mL 20.2 mL Volume HCI 20.4 mL 20.4 mL 22.2 mL Initial temperature solution 23.0°C 20.6°C 23.6 °C Final temperature 40.0 °C 37.2 °C 41.4 °C Part D Mass of calorimeter 49.441 g 48.532 g 49.058 g Mass of calorimeter + water 100.671 g | 102.874 g 96.915 g Initial temperature water 23.2 °C 24.4 °C 25.4 °C Mass of calorimeter + water + ice 107.485 g 110.007 g 104.923 g Final temperature 11.8 °C 13.0°C 12.0°C Theoretical values for data analysis Part A = 0.128 J/gºC Part B = -71.6 kJ/mol Part C = -57.3 kJ/mol Part D = 334 J/g 46.949 g 74.224 g 5.527 g 21.6 °C 41.4 °C

Answer 1

Here it should be noticed that heat capacity of calorimeter is given not specefic heat is given. So we dont have to multiple the mass of calorimeter.

PON solution for Past A for tried Here heat copacity of calonmeler = 22-1 5|C By using the Conservation energy prinople, caur know Heat goved by (abovendes tewater) - Heat test by legd, metal Nowo Heat gained by calenmatex = heat capacky x temporetowe change - 22.131°C X (240-21.006) = 22.1X3 J = 66.3 Joules Heat gained by water = mass of water te operofic heat of water x Hoapt charge = (1588 - 44:19)gx 4.1967/23 (24-21)°C = 896-0133 Jouks Heat lost by lead metal = mass x speafie heat coperty + tempt. change = 98-059 x x x (100 - 24) = (+6898.059) * Joules Now from equation ® unget e (48X98 - 059) &C = (896. 0133 + 663) Jodes = (896.0133 + 663 - 1 76898. 059 Similarly it can be calculated for foral 2 and 3.