1. Heating/Cooling Curves and Total Energy Change of a System Thermochemistry

2. Heating and Cooling Curves Shows how temperature and phase changes of a substance Y axis = temperature X axis = time

4. Heat Curve: Temperature vs. Time

5. Total energy change of a system When a given amount of energy is available to heat a substance, sometimes there is more then one change of state involved as the substance goes from an initial to a final temperature. This requires energy calculations for (1) changing the temperature in a certain state AND then (2) calculations for a change of state.

6. Total energy change of a system q total = q 1 + q 2 + q 3 …… To calculate each q you will use one of 2 equations that you already know!

7. Two choices: 1. To calculate the energy involved in a change of temperature of a substance without a change of state  q=mc ∆t (units J) 2. To calculate the energy involved in a change of state  q = nΔH (units kJ)

8. Steps: 1. Read question and sketch heating/cooling curve Label each q and important temperatures 2. Pick appropriate equations for each q 3. Solve for each q 4. Add all q’s together for total

9. Example: Calculate the total energy needed or released during the cooling of 1000 g of liquid iron (c= 0.82 J/g o C) to solid iron (c=0.52 J/g o C), where the initial temperature is 1700 o C to 80 o C.  The enthalpy of solidification for iron is -15 KJ/mol.  The melting point of iron is 1535 o C.

10. Heating Curve Temp (o C) Time (min) 1700 1535 80 q1 q2 q3

11. q total = q1 + q2 + q3 = mc Δt + n Δ H solid + mc Δt = [(1000g)(0.82 J/g C)(1535-1700)] + [(1000g/55.85 g/mol)(-15 KJ/mol) ] + [ (1000g)( 0.52 J/g C) (80-1535)] = -135300 J + (-268.6 KJ) + (-756600 J) **** Be sure to change J to KJ before adding******* = -1160 KJ

12. Example: How much energy is required to raise the temperature of 50.0 g of water (ice) at -15.0°C to a temperature of 175°C? Info:  c (H 2 O (l)) = 4.18 J/g C  c (H 2 O (s) and (g)) = 2.01 J/g C  ∆H melt = 6.02 kJ/mol  ∆H vap = 40.7 kJ/mol  Melting point = 0 o C  Boiling point = 100 o C First, start by drawing your own heat curve.

13. Five q’s for this question! the energy required to change the temperature of the ice from -15.0°C to 0°C, q 1 ; the energy involved in the change of state as the ice melts at 0°C, q 2 ; the energy involved to heat the liquid water from 0°C to 100°C, q 3 ; the energy involved in the change of state as the water boils, q 4 ; and finally the energy required to heat water vapour at 100°C to 175°C, q 5.

14. 100 o C 0 o C - 15 o C 175 o C Temp ( o C) Time (min)

15. q1 = mcΔt = 50.0 g x (2.06 J/g°C) x 15.0°C = 1540 J or 1.54 kJ q2 = nΔHfus = 50.0 g / (18.01 g/mol) x 6.03 kJ/mol = 16.7 kJ q3 = mcΔt = 50.0 g x (4.18 J/g°C) x 100.0°C = 20 900 J or 20.9 kJ q4 = nΔHvap = 50.0 g/(18.01 g/mol) x 40.8 kJ/mol = 113 kJ q5 = mcΔt = 50.0 g x (2.02 J/g°C) x 75.0°C = 7 580 J or 7.58 kJ qtotal = q1 + q2 + q3 + q4 + q5 = 1.54 kJ + 16.7 kJ + 20.9 kJ + 113 kJ + 7.58 kJ = 160 kJ

16. Practice: Pg 655 # 30-34 WS: Temperature and Phase Changes