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Changes in State

lev-stuart avatar
lev-stuart
10 diapositives · 191 Views · Oct 27, 2014

Changes in State. warm solid. warm gas. warm liquid. Temperature ->. Time ->. Warming Curve. 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. warm solid warm gas warm liquid Temperature -> Time -> Warming Curve 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. Calculations with Sp. 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 1. How many calories are released when 1.29 kg of water are cooled from 98.5°C to 20.0°C? Heat = (1 cal/g°C)(1,290 g)(78.5°C) = -101,000 cal This is negative because heat is removed.

  5. More Calculations with Sp. Heat 3. What is the final temperature when 20.0 cal of heat energy are added to 0.235 g of iron (specific heat 0.030 cal/g°C) at 19°C? Heat = (sp. heat)(mass)(T) 20.0 cal = (0.0300 cal/g°C)(T); 667 °C = T Heat is being added, so Tfinal = 19°C + 667 °C Tfinal = 686 °C

  6. 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.

  7. Calculations with Heats of Vap/Fusion 1. How much heat is needed to freeze 25.0 g of water? The heat of fusion of water is 1.14 kcal/mole. 25.0 gx (1 mole/18.0 g) x (1.14 kcal/1 mole) = 1.77 kcal 2. How much heat is needed to boil 125 g of alcohol? The heat of vaporization of alcohol is 24.5 cal/g. 125.0 gx x (24.5 cal/1 g) = 3060 cal

  8. 1 2 Temperature -> 3 4 5 Time -> Cooling Curve • Can you identify what’s happening at each stage of this graph? • Can you identify what states are present at each stage? • When is kinetic energy affected? When is potential energy affected?

  9. 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? First, cool the water to its freezing point. Heat = (1 cal/g°C)(35.0 g)(80.0-0.0°C) = - 2800 cal Second, freeze the water into ice. 35.0 g (1 mole/18.0 g)(1.14 kcal/1 mole) = 2.22 kcal Third, cool the ice to its final temperature. Heat = (0.50 cal/g°C)(35.0 g)(-25°C) = - 438 cal Last, add all the heats together. Make sure units agree! 2800 cal + 2,220 cal + 438 cal = -5460 cal

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