Change of Phase. What if…? If you were to remove the fast-moving particles from a substance, what would happen to the average speed of the remaining particles?

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Presentation transcript:

Change of Phase

What if…? If you were to remove the fast-moving particles from a substance, what would happen to the average speed of the remaining particles? If you were to remove the slow-moving particles, what would happen to the average speed of the remaining particles?

Evaporation When particles in a liquid gain enough energy to rip free from the bonds that hold them close to their neighbors, they change from liquid to gas. They evaporate.

Heat of Vaporization Scientists quantify the amount of energy required to evaporate with ‘the heat of vaporization’,  H vapor Substance Heat of vaporization (Joules /gram) Ammonia1,371 Butane320 Ethanol841 Water2,257 Aluminum10,897 It takes 2,257 J to change one gram of water to steam

Freezing When particles in a liquid lose so much energy that they are trapped by their neighbors, they change from liquid to solid. They freeze.

Heat of Fusion Scientists quantify the amount of energy released during freezing with ‘the heat of fusion’,  H fusion Substance Heat of fusion (Joules/gram) water334 methane58 acetone98 paraffin wax200– J are released when one gram of water changes to ice.

How much energy is required to increase the temperature of 1 gram of water from -10  C to 110  C?

From -10  C to 0  C Specific heat of ice, c ice = 2.03 J / g  C Change in temperature,  T = 10  C Mass = 1 g Q = cm  T = (2.03 J / g  C) (1 g) (10  C) = 20.3 J

From 0  C ice to 0  C water Heat of fusion for ice,  H fusion = J / g Mass, m = 1 g Q =  H fusion m = ( J / g) (1 g) = J

From 0  C to 100  C Specific heat of water, c water = J / g  C Mass = 1 g Change in temperature,  T = 100  C Q = cm  T = (4.186 J / g  C) (1 g) (100  C) = J

From 100  C water to 100  C steam Heat of vaporization for water,  H vap = 2,257 J / g Mass, m = 1 g Q =  H vap m = (2,257 J / g) (1 g) = 2,257 J

From 100  C to 110  C Specific heat of water vapor, c vapor = 1.86 J / g  C Change in temperature,  T = 10  C Mass = 1 g Q = cm  T = (1.86 J / g  C) (1 g) (10  C) = 18.6 J

Temperature Energy required -10  C to 0  C 20 J Ice to water 335 J 0  C to100  C 418 J Water to steam 2,257 J 100  C to 110  C 18 J

Temperature Energy required -10  C to 0  C 20 J Ice to water 335 J 0  C to100  C 418 J Water to steam 2,257 J 100  C to 110  C 18 J

How much energy is required to increase the temperature of 1 gram of water from -10  C to 110  C? From -10  C to 0  C20.3 J From ice to water333.5 J From 0  C to 100  C418.6 J From water to steam J From 100  C to 110  C18.6 J Total J