Chemistry Unit 8







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(Hf)

2.Heat of Vaporization(Hv)

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

Calculating Heat

q = m x ?T x Cp

q= heat(joules or calories)

m= mass(grams)

?T= change in temperature (Kelvin or ?Celsius)

Cp= specific heat of substance (amount of heat required to raise 1g of substance 1 ?Celsius)

Cp water = 4.18 J/gK or 1cal/gK (>1cal = 4.18 joules)

Calculate Heat (at phase changes)

1. Heat of fusion:melting/freezing

q= m x Hf

2.Heat of Vaporization:vaporizing/condensing

q= m x Hv

*?T is not used here because temperature remains constant

Phase Change Diagram


-antime the line goes up diagonally, use q= m?TCp

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

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

Examples – Lesson 1

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

400.0g(80.4?C)(4.18J/g?C)= 134429J!

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

q= m?TCp…m= q/?TCp

m= 3970J/(34?C)(.449J/g?C) = 260g!

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

q= m?TCp…substitue ?T to…q= m(Tf – Ti)Cp…[q/mCp] = (Tf Ti)…Tf = q/mCp + Ti (formula needed)

Tf = 2010calories/(202g)(1cal/g?C) + 70?C…10?C + 70?C = 80?C = Final Temperture!

*note: always round answer to nearest whole number!

Ex.4)2.5lb of water at 48?C 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

48?C + [15000J/(1136g)(4.18J/g?C)…48?C + 3.15?C = 51?C = Answer

Ex. 4)8750J of heat are applied to a piece of aluminum, causing a 56?C 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/(36?C)(.897J/g?C) = 174g!

Examples – Lesson 2

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

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

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

O?C to 100?C: q=46.0g(100?C)(4.18J/g?C)= 19228J

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

100?C to 114.0?C: q=46.0g(14?C)(2.02J/g?C)= 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/g?C. How much energy is required to raise the temperature of 10.0g of copper by 100?C?

q= 10.0g(100?C)(.0924cal/g?C)= 92.4 cal!

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

(boiling = vaporization)

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


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