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What is the melting point of this substance? 50˚C 100˚C The boiling point? Just to review before we start…

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Topic: Calculating Energy Changes at Phase Changes (H v and H f )

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It takes energy to heat stuff up! for pure substance in single phase - can calculate how much E needed using: for pure substance in single phase - can calculate how much E needed using: Q = mC T Q = mC T Q = energy in Joules Q = energy in Joules m = mass in grams m = mass in grams C = specific heat capacity C = specific heat capacity T = change in temperature = T f - T i T = change in temperature = T f - T i on other hand, when something cools down, energy is released! on other hand, when something cools down, energy is released!

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Temperature Time Q = mC liquid T Q = mC solid T Q = mC gas T C = specific heat capacity (amount heat required to raise temp of 1g of pure substance by 1 C) C is a physical constant unique for every pure substance CAN YOU FIND THE SPECIFIC HEAT OF WATER ON YOUR REFERENCE TABLE? IIIIII IV V

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Why can’t I use Q = mC T for II and IV?? Temperature Time IIIIII IV V Because T = 0, temp isn’t changing!!!!

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So, how do we calculate the amount of energy required during a phase change? H F = Heat of Fusion (Q = mH F ) H V = Heat of Vaporization (Q=mH V ) We use one of these two constants instead of specific heat and delta T Q = mC T

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H f = Heat of Fusion is amount energy required to change 1 gram is amount energy required to change 1 gram of pure substance from solid to liquid at its MP (meaning you aren’t changing the temperature) Is a physical constant Is a physical constant Check out Reference Table B, what is the heat of fusion for water? Check out Reference Table B, what is the heat of fusion for water? The Equation Q = mH f The Equation Q = mH f

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How much heat is absorbed when 10 grams of ice melts at 0 o C? Heat absorbed = Heat absorbed = mass of substance x heat of fusion of substance mass of substance x heat of fusion of substance Q = mH f = (10 g)(334 J/g) = 3340 J Q = mH f = (10 g)(334 J/g) = 3340 J Where does this energy go? Where does this energy go? Particles must overcome forces of attraction to move farther apart during phase change (s → l) Particles must overcome forces of attraction to move farther apart during phase change (s → l)

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H V = Heat of Vaporization is the amount energy required to convert 1 gram is the amount energy required to convert 1 gram of pure substance from liquid to gas at its BP (meaning you aren’t changing the temperature) Is a physical constant Is a physical constant Check out Reference Table B, what is the heat of vaporization for water? Check out Reference Table B, what is the heat of vaporization for water? The Equation Q = mH v The Equation Q = mH v

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How much energy does it take to vaporize 10 g of water? Q = mH v Q = mH v Q = (10 g)(2260 J/g) = 22600 J Q = (10 g)(2260 J/g) = 22600 J

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H 2 O changing from solid to liquid requires 3,340J/g It takes a lot more energy to go from liquid to gas than from solid to liquid. Why? H 2 O changing from liquid to gas requires 22,600J/g * greater energy required to change from liquid to gas because particles are spreading farther apart!

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Temperature Time Q = mC solid T IIIIII IV V Q = mH F Q = mH V Q = mC LIQUID TQ = mC gas T

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Heating curve of H 2 O

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3 equations for Q 1. Q = mC T 2. Q = mH f 3. Q = mH v figure out which to use figure out which to use depends on section of heating curve depends on section of heating curve look for hints in word problem look for hints in word problem

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Q = mC T Temperature changed Temperature changed T ↑ T ↑ T ↓ T ↓ Initial temperature Initial temperature Start temperature Start temperature Final temperature Final temperature Ending temperature Ending temperature From __ ˚C to __ ˚C From __ ˚C to __ ˚C Water Water

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Q = mH f Ice Ice Freezing Freezing Melting Melting Occurs at 0 C (for H 2 O) Occurs at 0 C (for H 2 O) At constant temperature At constant temperature

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Q = mH v Steam Steam Boiling Boiling Condensation Condensation Occurs at 100 C (for H 2 O) Occurs at 100 C (for H 2 O) At constant temperature At constant temperature

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heating rate = 150 J/min If the substance takes 4 minutes to melt, how much heat energy was used to melt it? 150J/min x 4min = 600J

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