The Fundamentals of Calorimetry
Insert graph of the data from Data Table 1, plotting temperature vs. time, and use the Y intercept to find the temperature at time 0 when the two volumes of water are mixed.
|Trial 1 (include units)||Trial 2 (include units)||Calculations (for Trial 1)|
|Initial temperature of cold water|
|Initial temperature of warm water|
|Temperature at time 0 from graph, T0|
|Heat lost by hot water|
|Heat gained by cold water|
|Heat gained by calorimeter|
|Temperature change of calorimeter|
|Heat capacity of calorimeter|
|Average Heat capacity of calorimeter|
Report data with correct units. Show calculations below the table
|Mass of water
|Mass of salt
|Moles of salt
|Change in Temperature|
|Heat released/absorbed by water|
|Heat released/absorbed by the calorimeter|
|Enthalpy of solution
|Molar Enthalpy of solution|
|Average molar enthalpy of solution|
Show calculations for first column:
Moles of salt
Heat released/absorbed by the solution
Heat released/absorbed by the calorimeter
Enthalpy of solution (J)
Enthalpy of solution (kJ)
Molar Enthalpy of solution
Average molar enthalpy of solution
Insert graphs of ΔT vs. grams of salt for NH4Cl
What sort of relationship exists between the temperature change and the mass of the salt dissolved? Explain your answer.
The actual molar enthalpy of solution for calcium chloride is -81.3 kJ/mol, whereas the molar enthalpy of solution of ammonium chloride is 14.8 kJ/mol. Calculate the percent error of your results. Show the calculations.
Use the value of enthalpy you determined for CaCl2 to calculate what quantity of salt will be needed to make a chemical hot pack. Assume using 100 g (100 mL) of water and changing the temperature from 25 °C to 60 °C. Show all calculations.
Use the graph you drew for NH4Cl to determine the amount of salt needed to make a chemical cold pack. Assume using 100 g (100 mL) of water. The temperature should go down to 3.0 °C from a room temperature of 25 °C. Show all calculations.
Suggest practical ways in which the calorimeter or lab protocol could be improved to decrease percent errors.
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