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Do now

Do now. 1) Take out graphic organizers (turn in any late work) 2) Work on the following problem 55 grams of Compound X has 50 kJ of energy before decomposing into elements Y and Z, which have a combined energy of 20 kJ. What is the combined mass of elements Y and Z?

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Do now

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  1. Do now 1) Take out graphic organizers (turn in any late work) 2) Work on the following problem 55 grams of Compound X has 50 kJ of energy before decomposing into elements Y and Z, which have a combined energy of 20 kJ. • What is the combined mass of elements Y and Z? • Physical or chemical change? • Is this an exothermic or endothermic reaction? • What is the ΔH (heat of reaction)?

  2. DRESS APPROPRIATELY! • We will be doing labs pretty frequently this next week so wear proper attire (hair tied back, no open toed shoes, etc…) • If you do not have your safety forms AND your safety lab turned in, you will not be allowed to do the labs…

  3. Calorimetry; Measuring heat flow • Ever wonder how they find out how many calories are in your food? • The answer; Calorimetry Calo-what?

  4. Calorimetry; Measuring heat flow • When a reaction occurs, potential energy is either gained (+ ΔH) or lost (- ΔH) • Can’t use a thermometer to quantify and measure this, so we need to use something else…

  5. Things we know • 1 gram of water needs 4.18 Joules of energy to increase its temperature 1 degree celsius • we call this its specific heat • 4.18 joules / (grams ocelsius) • During a reaction, energy (joules) is going to either be gained from or lost to the surroundings • How can we use this to figure out the heat change?

  6. What about this? • Get the stuff to react • Energy could be absorbed by water… • The temperature change and amount of water would tell us how much energy was absorbed…

  7. Calorimetry • What will we need to know to figure out heat exchange? • How much energy is needed to change water’s temperature  4.18 J/(g*degree celsius) • Mass of water • Change in temperature (ΔT) • Put it together…

  8. Formula • J / (grams * ocelsius) • How can we get just joules? • Need grams and degree celsiusto cancel out… • Multiply by grams and degree celsius!

  9. Where the equation came from • J/(g* ocelsius) * g * ocelsius J • Rearrange • Joules = g * J/(g*ocelsius) * ocelsius • Heat change = mass * specific heat * Δ temp • q = m * C * ΔT • Wishing there was a song to remember all of this??? • YOU GOT IT!

  10. An example • How many joules are absorbed by 100.0 grams of water if the temperature is increased from 35.0oCto 50.0 oC? • q = m c ΔT • q= 100.0 g * 4.18 J/(goC) * 15.0oC • q= 6270 Joules

  11. What if we had to solve for the mass of water? • A sample of water is heated by 20.0 oC by the addition of 80.0 J of energy. What is the mass of the water? • Rearrange the formula to solve for m… • q = m c ΔT………Divide each side by c ΔT • q / (c ΔT) = (m c ΔT) / (c ΔT) • m= q / (c ΔT)

  12. m= q / (c ΔT) • A sample of water is heated by 20.0 oCby the addition of 80.0 J of energy. What is the mass of the water? • m= 80.0 J / (4.18 J/(goC) * 20.0oC) • m= 0.956937799 grams • m= 0.957 g

  13. How about a tough one… • 200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at 25.000 oC.What will the final temperature be? • Hmm…wants final temperature and tells us the initial… • Need to calculate ΔT… then we need to rearrange the formula • Δ T = q / m C

  14. Δ T = q / m C • 200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at 25.000 oC. What will the final temperature be? • Δ T = q / m C • Δ T = 200. J / [80.0* 4.18 J/(goC)] • Δ T = 0.598086124oC • Δ T = 0.598oC • NOT DONE YET!!!

  15. Δ T = 0.598oC • 200. J of energy is absorbed by an 80.0 g sample of water in a calorimeter at 25.000 oC. What will the final temperature be? • Says energy is absorbed by the water, so the temperature is going to…increase • Tfinal = Tinitial + Δ T • Tfinal = 25.000 oC + 0.598oC • Tfinal = 25.598 oC

  16. Reference Table I • ‘Heats of reaction’; known heats of reaction are listed • C(s) + O2(g)  CO2(g) ∆H= –393.5 kJ • Is this exothermic or endothermic? • Exothermic

  17. C(s) + O2 (g) CO2 (g) ∆H= –393.5 kJ • What if the reverse reaction happened, what would the heat of reaction be? • CO2 (g)C(s) + O2 (g) ∆H=??? • If it is – 393.5 one way…then it would be the opposite the other way • ∆H= +393.5 kJ • Exothermic or endothermic? • Endothermic!

  18. With your neighbor • Work on the calorimetry homework

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