1. Experiment Find the Enthalpy change of Dissolution

2. Starter What is the enthalpy change in Jmol -1 when 5g of sodium hydroxide is added to 100g of water. Initial temp of water 20 o C Final temp of water 40 o C Specific heat capacity of water = 4.18 kJ kg -1 K - 1

3. Calculation Energy change of water = m.c.∆T Conservation of energy says that thermal energy lost = thermal energy gained. Therefore ∆H = m.c.∆T Determine the number of moles of sodium hydroxide ∆H mol -1 = ∆H / moles

4. Plenary: Uncertainty Perform the same calculation as in the starter but this time use the following uncertainties. 5.000 ± 0.0005g of sodium hydroxide 100.000 ± 0.0005 g of water. Initial temp of water 20.0 ± 0.05 o C Final temp of water 40.0 ± 0.05 o C Specific heat capacity of water = 4.183 kJ kg -1 K -1

5. Uncertainty of Reference Values Specific heat capacity of water = 4.183 kJ kg -1 K -1 Assume precision is limited to the number of decimal places i.e 4.183 ± 0.0005 kJ kg -1 K - 1

6. Calculation of Uncertainty ∆T = (40.0 ± 0.05 o C) - (20.0 ± 0.05 o C) = 20.0 ± 0.1 o C % Uncertainty = (0.1 / 20) x 100 = 0.5%

7. The Trick Sometimes the % uncertainty of one is much larger than all the other % uncertainties just use this and ignore the others. i.e ∆H = m.c.∆T = 100 x 4.183 x 20 = 8366 kJ kg -1 K - 1 Uncertainty = 0.5 /100 x 8366 = 42 Answer 8366 ± 42 kJ kg -1 K - 1