heat of Neutralization experiment 28 Heat Capacity of Calorimeter Temp of calori
ID: 907209 • Letter: H
Question
heat of Neutralization experiment 28 Heat Capacity of Calorimeter Temp of calorimeter and water before mixing degree C Temp of warm water degree C Maximum temp. determined from your curve degree C Heat lost by warm water (temp decrease x 50.0g x 4.184 J/K.g) = J 5. Heat gained by cooler water (temp increase x 50.0g x 4.184 J/K.g) = J 6. Heat gained by the calorimeter [(4) - (5)] = J 7. Heat capacity of calorimeter: heat gained by the calorimeter / temperature increase J/K 8. Heat of Neutralization of HCl-NaOH degree C 1. Temp. of calorimeter and NaOH Temp. of HCl Delta T determined from your curve after adding HCl to the NaOH degree C Heat gained by solution (temperature increase x 100 g x 4.184 J/K-g) = J Heat gained by calorimeter (temperature increase x heat capacity of calorimeter) = J Total joules released by reaction [(3) + (4)] = JExplanation / Answer
As per the given concept we need to calculate the heat lost by the warm water and also we need to to calculate the heat gain by the cold water .
Since:
temperature of the warm water = 42.0 oC.
maximum temperature as per the curve = 33.0 oC.
mass = 50.0 g.
specific heat = 4.184 J/K-g
As we know the formula for
heat lost by warm water = temperature decrease x mass x specific heat.
Now substitute the values according to the formula.
heat lost by warm water = (42.0-33.0) x 50.0 x 4.184
= 9 x 50.0 x 4.184
= 1882.8 J
temperature of the water = 22.8 oC.
maximum temperature as per the curve = 33.0 oC.
mass = 50.0 g.
specific heat = 4.184 J/K-g
As we know the formula for
heat gained by cool water = temperature increase x mass x specific heat.
Now substitute the values according to the formula.
heat gained by cool water = (33.0-22.8) x 50.0 x 4.184
= 10.2 x 50.0 x 4.184
= 2133.84 J
heat gained by the calorimeter = heat lost by warm water - heat gained by cool water
= 2133.84 - 1882.8
= 251.04 J
Heat capacity of calorimeter = heat gained by the calorimeter/ temperature increase
= 251.04 J/10.2 oC
= 209.2 J/K.
= 24.61 J/K .
HCl (aq) + NaOH (aq) NaCl (aq) + H2O (l)
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