1. Warmup Given the following equations: H 3 BO 3(aq)  HBO 2(aq) + H 2 O (l) ΔH rxn = -0.02 kJ H 2 B 4 O 7(aq) + H 2 O (l)  4HBO 2(aq) ΔH rxn = -11.3 kJ H 2 B 4 O 7(aq)  2B 2 O 3(s) + H 2 O (l) ΔH rxn = 17.5 kJ find the Δ H for this overall reaction: 2H 3 BO 3(aq)  B 2 O 3(s) + 3H 2 O (l) Answer: 14.4 kJ

2. Specific Heat Capacity and Latent Heat

3. Every substance has its own specific heat capacity because substances respond differently to being heated A molecule of water can absorb much more heat than an atom of metal without increasing in temperature. Water can absorb 4.180 J per gram of water. That 1g water will increase by 1⁰C. This is it’s specific heat capacity(c).

4. q = m c ∆T q stands for c stands for m stands for ∆T stands for heat energy (J) specific heat (J/g°C) mass (g) temperature change (°C) Specific heat capacity: the amount of energy required to raise the temperature of one gram of a substance by one degree celsius.

5. Ex 1: A 21 g cube of silver is heated from 30. 0 C to 46 0 C. How much heat did the silver absorb? (c silver = 0.240 J / g °C) q c m ∆T q = c x m x ∆T ∆T = T f - T i ? 0.240 J/g  C 21 g 16 °C q = 0.240 J/g  C x 21 g x 16 °C q = 81 J or 0.081 kJ

6. Ex 2: A 21.4 g sample of gold absorbs 152.8 J of energy when its temperature is raised from 23.5 0 C to 78.9 C. Find the specific heat of gold. q c m ∆T q = c x m x ∆T ? 152.8 J 21.4 g 55.4 °C c = 0.129 J/g  C

7. Ex 3: If 600. Joules are applied to 250. grams of liquid water at 16.000 0 C, what will be the new temperature of the water? (c water = 4.18 J/g °C) q c m ∆T q = c x m x ∆T ? 600. J 250. g 4.18 J/g °C ∆T = 0.574  C T f = 16.574  C

8. Ex 4: The temperature of a cup of hot tea is 50.°C and normal body temp is 37°C. If you drink a 200.0g cup of tea, how much heat energy transfers to your body? Assume your whole body temperature increases by 0.1 °C heat lost by = heat gained drink by body -q d = q b -mc∆T = q b -(200.0g)(4.180J/g  C)(-12.9  C) = q b q b = 10784 J….round… Your body will absorb 11,000 J

9. Complete the diagrams below with the appropriate phase change names: endothermic phase changes require (absorb) energy from the surroundings exothermic phase changes release (give off) energy to the surroundings deposition freezing sublimation condensatio n vaporization melting

10. What do you notice about the diagram below? During a phase change, there is NO ΔT. All heat applied during a phase change is used to break the intermolecular forces that keep the molecules together “Latent heat” is hidden heat…no change in temperature. water and steam ice and water

11. Ex 1: How much heat must be added to a 25g ice cube at 0ºC to completely melt it to water at 0ºC? Heat of fusion: amount of energy required to completely melt 1 gram of a substance (freezing would use the same constant, but negative). q = m∆H f q = (25g)(334 J/g) 8400 J Important Information for water: ∆H f = 334 J/g ∆H v = 2260 J/g C ice = 2.06 J/g ∙°C C liquid = 4.18 J/g ∙°C C vapor = 1.87 J/g ∙°C You’ll be given all this stuff on the test!

12. Ex 2: How much heat must be added to a 25g sample of liquid water at 100ºC to completely vaporize it to water into steam at 100ºC? Heat of vaporization: amount of energy required to completely vaporize 1 gram of a substance (condensation would use the same constant, but negative). q = m∆H v q = (25g)(2260 J/g) q = 57000 J Important Information for water: ∆H f = 334 J/g ∆H v = 2260 J/g C ice = 2.06 J/g ∙°C C liquid = 4.18 J/g ∙°C C vapor = 1.87 J/g ∙°C

13. Ex 3: How much energy is lost when 20.0 g of water decreases from 303.0 °C to 283.0 °C? q = mcΔT = (20.0g)(1.87 J/g°C)(-20.0°C) q = - 748 J or, 748J are lost Important Information for water: ∆H f = 334 J/g ∆H v = 2260 J/g C ice = 2.06 J/g ∙°C C liquid = 4.18 J/g ∙°C C vapor = 1.87 J/g ∙°C

14. Ex 4: What is the total change in energy when 100.0g steam at 100.ºC condenses to water at 80. ºC? q = - m ∆H v q = - (100.0g)(2260 J/g) q = mc∆T q = (100.0g)(4.18J/g ° C)(-20. ° ) q total = -234360 J or - 2.3 x 10 5 J Important Information for water: ∆H f = 334 J/g ∆H v = 2260 J/g C ice = 2.06 J/g ∙°C C liquid = 4.18 J/g ∙°C C vapor = 1.87 J/g ∙°C = - 226000 J = - 8360 J

15. Ex 5: What is the TOTAL amount of heat necessary to bring 15.0g ice at -15.0⁰C up to steam at 135 ⁰ C? q = mcΔT = (15.0g)(2.06 J/g∙°C)(15.0 ⁰ C) = 464 J q = m∆H f = (15.0)(334 ) = 5010 J q = m∆H v = (15.0)(2260) = 33900 J q = mcΔT =(15.0)(1.87)(35.0) = 982 J q = mcΔT =(15.0)(4.18)(100.0) = 6270 J q TOTAL = 46626 J 4.66 x 10 4 J