1.Melting

2.Freezing

3.Vaporization

4.Condensation

5.Sublimation

6.Deposition

1.solid to liquid

2.liquid to solid

3.liquid to gas - different from evaporation.

(Vaporization is when you ADD heat to convert)

4.gas to liquid

5.solid to gas

6.gas to solid

the total energy of molecules. Depends on amount of substance you have

i.e.)a big tub of boiling water has more "heat" than a cup of boiling water, though they're the same temperature

Energy and Change of State

(Solid to Liquid to Gas)

1.Heat of Fusion(H_f)

2.Heat of Vaporization(H_v)

1.the amount of heat required to melt/freeze one gram of substance at its melting or freezing point

2.the amount of heat required to evaporate one gram of substance at its boiling/vaporizing or condensing point

another word for "heat" - the enthalpy of fusion or vaporization

q = m x ΔT x C_p

q= heat(joules or calories)

m= mass(grams)

ΔT= change in temperature (Kelvin or ºCelsius)

C_p= specific heat of substance (amount of heat required to raise 1g of substance 1 ºCelsius)

C_p water = 4.18 J/gK or 1cal/gK (→1cal = 4.18 joules)

1. Heat of fusion:melting/freezing

q= m x H_f

2.Heat of Vaporization:vaporizing/condensing

q= m x H_v

*ΔT is not used here because temperature remains constant

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-antime the line goes up diagonally, use q= mΔTC_p

-the horizontal line that changes from solid to liquid, use q= mH_f

-the horizontal line that changes from liquid to gas, use q=mH_v

Ex. 1)How much heat is required to raise 400.0grams of water from 11.3ºC to 91.7ºC?

400.0g(80.4ºC)(4.18J/gºC)= 134429J!

Ex. 2)What mass of iron would increase 34ºC if 3970 joules of energy is added?

q= mΔTC_p...m= q/ΔTC_p

m= 3970J/(34ºC)(.449J/gºC) = 260g!

Ex. 3)If 202 grams of water at 70.0ºC gains 2010 calories of heat, what will its FINAL temperature (T_f) be?

q= mΔTC_p...substitue ΔT to...q= m(T_f - T_i)C_p...[q/mC_p] = (T_f - T_i)...T_f = q/mCp + T_i (formula needed)

T_f = 2010calories/(202g)(1cal/gºC) + 70ºC...10ºC + 70ºC = 80ºC = Final Temperture!

*note: always round answer to nearest whole number!

Ex.4)2.5lb of water at 48ºC will be at what temperature after 15,000 joules of heat are added?

convert lb to grams...2.5lb x (1kg/2.2lb) x (1000g/1kg) = 1.136grams

48ºC + [15000J/(1136g)(4.18J/gºC)...48ºC + 3.15ºC = 51ºC = Answer

Ex. 4)8750J of heat are applied to a piece of aluminum, causing a 56ºC increase in its temperature. What is the mass of the aluminum?

(look up in chemistry reference table to find the specific heat of aluminum)

8750J/(36ºC)(.897J/gºC) = 174g!

Ex. 1)If you have a 46.0gram sample of H2O at a temperature of -58.0ºC. How many joules of energy are necessary to convert the sample of ice to steam at 114.0ºC?

-58.0ºC to 0ºC: q=46.0g(58ºC)(2.05J/gºC)= 5469.4J

melt: q=46.0g(334J/g)= 15364J

OºC to 100ºC: q=46.0g(100ºC)(4.18J/gºC)= 19228J

Vaporize: q=46.0g(2260J/g)= 103960J

100ºC to 114.0ºC: q=46.0g(14ºC)(2.02J/gºC)= 1300.88J

...add all the values up to get: 145322J!

Ex. 2)How much energy will be needed to melt 4.24 grams of Pd? The enthalpy of fusion for Pd is 162J/g.

q= 4.24g(162J/g) = 687J!

Ex. 3)The specific heat of copper is .0924cal/gºC. How much energy is required to raise the temperature of 10.0g of copper by 100ºC?

q= 10.0g(100ºC)(.0924cal/gºC)= 92.4 cal!

Ex. 4) How many joules are required to boil 150g of water?

(boiling = vaporization)

q= 150g(2.260J/g)= 339000J!