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Changes in State

Changes in State. warming the solid. warming the gas. warming the liquid. Temperature ->. Time ->. Warming Curve. Consider what happens to an ice cube which is heated. boiling. melting. warm gas. Temperature ->. warm liquid. warm solid. Time ->. Within One State.

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Changes in State

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  1. Changes in State

  2. warming the solid warming the gas warming the liquid Temperature -> Time -> Warming Curve Consider what happens to an ice cube which is heated. boiling melting

  3. warm gas Temperature -> warm liquid warm solid Time -> Within One State Risingtemperature indicates a change in kinetic energy: molecules are moving faster. This is measured by the specific heat (heat capacity): the calories needed to raise the temperature of 1 gram of a substance by 1°C.

  4. boil Temperature -> melt Time -> Changing States A temperatureplateau indicates a change in potential energy: bonds between molecules are changing, and energy can be stored in them. This is measured by the heat of vaporization (gas/liquid) orheat of fusion(solid/liquid): the calories needed to change the state of 1 gram.

  5. SUMMARY Temperature -> Time -> warm gas boil sp. heat gas warm liquid heat of vaporatization melt sp. heat liquid warm solid heat of fusion sp. heat solid

  6. cal or kcal Joules or kJ grams or kg moles Units You will see a variety of units for both energy and amount of material in these problems. Energy Amount

  7. 1 2 Temperature -> 3 4 5 Time -> Cooling Curve • W happening at each stage of this graph? • What states are present at each stage? • What property describes each stage? • When is kinetic energy affected? When is potential energy affected? Answers are in the notes.

  8. Calculations with Specific Heat 1. How many calories are needed to warm up 3.45 g of water from 18.5°C to 60.0°C? The specific heat of water is 1 cal/g°C. Heat = (1 cal/g°C)(3.45 g)(41.5°C) = 143 cal 2. How many calories are removed (released) when 1.29 kg of H2O are cooled from 98.5°C to 20.0°C? Heat = (1 cal/g°C)(1,290 g)(78.5°C) = 101,000 cal Technically, this is negative because heat is removed. We will not be using this convention, leaving all energy without signs and using works to decribe addition/loss of energy.

  9. More Calculations with Specific Heat 3. What is the final temperature when 85.0 J of heat energy are added to 23.5 g of iron (specific heat 0.125 J/g°C) at 19°C? (0.125 J/g°C)* (23.5 g) * (T) = 85.0 J T = 28.9 °C Heat is being added, so Tfinal = 19°C + 28.9°C Tfinal = 47.9 °C Note: If the problem stated that heat was being removed the iron, then T would still be 28.9 °C but the final temperature would be -9.9 °C. Do you see why?

  10. 1.14 kcal = x kcal alternative proportional solution 18.0 g 25.0 g x = 1.77 kcal 102.4 J = x J alternative proportional solution 1 g 125 gx = 12,800 J Heats of Vaporization & Fusion (1.14 kcal/1 mole) x [25.0 gx (1 mole/18.0 g)] = 1.77 kcal 1. How much heat must be released to freeze 25.0 g of water? The heat of fusion of water is 1.14 kcal/mole. 2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 102.4 J/g. (102.4J/1 g) x 125 g = 12,800 J

  11. Heat of fusion H2O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C 100 °C 0 °C Putting it together How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? FIRST: Sketch a heating or cooling curve and mark the plateaus and your starting/ending points.  starts at 80 °C; liquid Temperature ->  ends at -25 °C; solid This tells you there are 3 stages to consider: (a) cooling the liquid, (b) freezing, (c) cooling the solid. Time ->

  12. Heat of fusion H2O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C Putting it together How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? Second: Find the properties that control each stage of the problem. cooling the liquid = specific heat of liquid: 1 cal/g°C freezing = heat of fusion: 1.14 kcal/mol cooling the solid = specific heat of solid: 0.50 cal/g°C The units of each property show you how to do the math!

  13. How much heat energy must be removed from 35.0 g of water at 80.°C to turn it into ice at -25 °C? Heat of fusion H2O = 1.14 kcal/mol Sp. Heat of Ice = 0.50 cal/g°C Putting it together 1. Cool the water to its freezing point. (1 cal/g°C)(35.0 g)(80.-0.0°C) = 2800 cal 2. Freeze the water into ice. (1.14 kcal/1 mole) (35.0 g)(1 mole/18.0 g) = 2.22 kcal 3. Cool the ice to its final temperature. (0.50 cal/g°C)(35.0 g)(0-[-25°C]) = 440 cal Last, add all the heats together. Make sure units agree! 2800 cal + 2220 cal + 440 cal = 5460 cal

  14. Entropy crystalline solid highly ordered minimum entropy liquid some order some entropy gas very random maximum entropy

  15. cal/g, cal/mole kcal/g, kcal/mole J/g, J/mole cal/g°C J/g°C

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