The heat of capacity is usually a positive sign and hence the value for the heat capacity of the calorimeter can be given as 23.4 ± 0.1 J/OC.
Sample for the Calculations Using Trial #1
The weight of cold water was calculated by getting the difference between the weight of the calorimeter together with cold water and the weight of calorimeter as follows
Weight of cold water=51.7g-3.3g
=48.4 grams
The weight of hot water was calculated as follows
=Final weight of calorimeter + water -Weight of calorimeter + cold water
=100.4 g-51.7 g
=48.7 grams
The heat lost by hot water was calculated as follows
heat lost=heat capacity of water ×change in temperature×wight of hot water
=4.18 Jg-1℃-1×48.7 g×7.3 ℃
=1486.03 J
Heat gained by cold water was calculated as follows
heat gained=heat capacity of water ×change in temperature×wight of hot water
=4.18 Jg-1℃-1×48.4 g×8.2 ℃
=1658.96 J
Heat lost to calorimeter was calculated as follows
heat lost to calorimeter=heat lost by hot water-heat gained by cold water
=1486.03 J-1658.96 J
=-172.93 J
The Heat Capacity of Calorimeter was calculated as follows
=Heat lost tocalorimeterTemperature change
=172.93J8.2℃
=-21.088 J/℃
The molar heat of neutralization is therefore 59.7 ±0.1 kJ/mole
Sample for the Calculations Using Trial #1
Heat absorbed by 2.5 M NaCl solution was calculated as follows
=specific heat capacity of the solution×Weight×Temperature change
3.59 Jg-1℃-1×109.2 g×36.1 ℃
=14152.211 J
Heat absorbed by calorimeter was calculated as follows
=heat capacity of the calorimeter×Temperature change
23.44 J℃-1×36.1 ℃
=846.214 J
Total heat of neutralization was calculated as follows
=heat absorbed by NaCl solution+Heat absorbed by calorimeter
=14152.211 J+846.214 J
=14998.425 J
Heat of neutralization per mole of water was calculated as follows
=total heat of neutralizationmoles of water produced
=14998.425 J0.250 moles×1000
=59.994 kJ/mole
The molar heat of hydration is therefore 111.6 ±0.1 kJ/mole
Sample for the Calculations Using Trial #1
The heat absorbed by dilute acid solution was calculated as follows
=specific heat capacity of the solution×weight×temperature change
=4.12 Jg-1℃-1×99 g×37.8 ℃
=15417.86 J
The heat absorbed by the calorimeter was calculated as follows
=heat capacity of the calorimeter×weight×temperature change
=23.44 J℃-1×37.8 ℃
=886.064 J
The total heat evolved in the hydration process was calculated as follows
=heat absorbed by the dilute acid solution+Heat absorbed by calorimeter
=15417.864 J+886.064 J
=16303.928 J
The heat of neutralization per mole was calculated as follows
total heat evolved in the hydration process moles of H2SO4
=16303.928 J0.144 moles×1000
=113.222 kJ/mole
The molar heat of solution is therefore -20.0 ±0.1kJ/mole
Sample for the Calculations Using Trial #1
The heat lost by ammonium chloride solution was calculated as follows
=specific heat capacity of the solution×weight×temperature change
=3.59 Jg-1℃-1×99 g×-12.9 ℃
=-4584.789 J
The heat lost by the calorimeter was calculated as follows
=heat capacity of the calorimeter×weight×temperature change
=23.44 J℃-1×-12.9 ℃
=-302.387 J
The total heat absorbed in the solution reaction was calculated as follows
=heat lost by the ammonium chloride solution+Heat lost by calorimeter
=-4584.789 J+-302.387 J
=-4887.176 J
The heat of solution per mole was calculated as follows
total heatabsorbed in the solution reaction moles of NH4Cl
=-4887.175 J0.22 moles×1000
=-22.21 kJ/mole
