1. Determining Enthalpy of Rxn by Experiment
To measure enthalpy changes the system must be isolated (separated) from the surroundings
The heat change is measured by the temperature change of the system
The temperature change of any substance is related to its Specific Heat Capacity

2. Specific Heat Capacity (c)
Amount of energy needed to raise the temperature of one gram of substance by 1°C (or 1°K)
Expressed in units of J/g°C
Substance
Specific Heat Capacity J/g°C
Aluminum
0.900
Ethanol
2.46
Water (liquid)
4.184
Air
1.02
Wood
1.76

3. Using Specific Heat Capacity
Can be used to:
Calculate the amount of energy needed to heat a given mass by a certain number of degrees.
Calculate the amount of heat released when the temperature of a given mass decreases by a certain number of degrees.

4. Specific Heat Capacity vs. Heat Capacity (C)
Specific heat capacity is constant for all samples of the same substance
Eg J/g°C for all liquid water
Heat capacity – relates the heat of a specific sample or system to a change in temperature
Units are kJ/°C
Eg. Comparing a cup of water to a bath tub
Both samples have the same specific heat capacity
But the bath tub has a much higher heat capacity
It takes much more heat to raise the temperature of the water in the bath tub by the same amount.

5. Specific Heat Transfer and Heat Transfer
Using the following equation we can calculate:
Heat change of a substance relative to the mass
The specific heat capacity
The change in temperature
Q = mc∆T where, Q = heat (J)
m = mass (g)
c = specific heat capacity (J/g°C)
∆T = Tf (final) – Ti(initial) (°C or °K)

6. Example: Calculate the amount of energy needed to heat 100 g of water from 20.0°C to 45.0°C
m = 100 g ∆T = 45.0°C – 20.0°C = 25.0°C
C = J/g°C
Q = mc∆T
Q = (100 g)(4.184 J/g°C)(25.0°C) = J = 1.05 x 104 J
Therefore, 1.05 x 104 J of energy is needed to raise the temperature of 100 g of water by 25°C

7. Practice problems Pg. 302# 9, 10, 11, 12, 13

8. Measuring Heat Transfer
Calorimeters – used to measure the enthalpy changes for chemical and physical reactions.
Insulate the system from the surroundings
The temperature change of the system allows you to determine the heat released or absorbed by the reaction
For an exothermic reaction
Qreaction = -Qinsulated system

9. Sample Problem – Enthalpy of a chemical reaction
CuSO4(aq) + NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq) Volume of CuSO4 = 50.0 ml Volume of NaOH = 50.0 ml Initial temperature, Ti = 21.4°C Final temperature, Tf = 24.6°C Assume, specific heat capacity = water = J/g°C

10. Steps to solve
Determine total volume of reactants. Determine mass of reactants assuming density = 1 g/ml (density of water)
Use Q = mc∆T to calculate heat absorbed
Use Qreaction = -Qsolution to determine heat released
Determine number of moles for each reactant. If necessary determine limiting reactant. Use the limiting reactant to get ∆H of the reaction (kJ/mol)
Use ∆H to write a thermochemical equation

11. Solution The total volume is 50.0 ml + 50.0 ml = 100ml
Assuming a density of 1.00 g/ml (same as water)
m = DV = 1.00g/ml(100ml) = 100 g
2. Amount of heat absorbed by solution
Qsolution = mc∆T
= (100 g)(4.184 J/g°C)(24.6 °C – 21.4 °C)
= 1300 J
3. Heat change for the reaction is J

12. Solution Con’t 4. Calculate number of moles of CuSO4
300 mol/L x L = mol
Calculate number of moles of NaOH
0.600 mol/L x L = mol
The reactants are present in a 2:1 ratio…the same as the balanced reaction, therefore, no limiting reagent.

13. Solution Continued ∆H = heat released/number of moles
∆H of the reaction in kJ/mol CuSO4
∆H = heat released/number of moles
= J/ mol CuSO4
= J/mol CuSO4
= -89 kJ/mol CuSO4
Therefore the enthalpy change is -89 kJ/mol CuSO4
The thermochemical equation is:
CuSO4(aq) + NaOH(aq)  Cu(OH)2(s) + Na2SO4(aq) ∆H = -89 kJ