Calculating Energy Changes at Phase Changes

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Presentation transcript:

Calculating Energy Changes at Phase Changes Q = mHf Q = mHv

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

What is the melting point of this substance? The boiling point?

Heat of Fusion = amount energy required to change 1 gram of pure substance from solid to liquid at its mp Heat of Fusion = Hf = physical constant Hf for water = 334 J/g (Table B)

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

Heat of Vaporization = amount energy required to convert 1 gram of pure substance from liquid to gas at its bp Heat of vaporization = Hv = physical constant Hv for water = 2260 J/g

How much energy does it take to vaporize 10 g of water? Q = mHv Q = (10 g)(2260 J/g) = 22600 J It takes a lot more energy to go from liquid to gas (22,600J) than from solid to liquid (3,340J). Why?

H2O changing from solid to liquid requires 3,340J/g

H2O changing from liquid to gas requires 22,600J/g greater energy required to change from liquid to gas because particles must spread farther apart!

Heats of fusion & vaporization Determined in calorimetry experiments

Temperature Time Q = mHv Q = mHf Q = mCliquidT Q = mCsolidT Q = mCgasT Time

3 equations for Q 1. Q = mcT 2. Q = mHf 3. Q = mHv figure out which to use depends on section of heating curve look for hints in word problem

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

Q = mHf Ice Freezing Melting Occurs at 0C (for H2O) At constant temperature

Q = mHv Steam Boiling Condensation Occurs at 100C (for H2O) At constant temperature