Specific Heat Specific Heat CapacityCapacity

Imagine…Imagine…A hot day in Arizona…in your back A hot day in Arizona…in your back

yard is a yard is a metal barbequemetal barbeque and a and a glass of waterglass of water. Would you want to . Would you want to stick your hand in the water or stick your hand in the water or touch the bbq?touch the bbq?

You would want to touch the You would want to touch the waterwater because different substances have because different substances have different abilities to absorb heat!!different abilities to absorb heat!!

Specific Heat CapacitySpecific Heat CapacityThe quantity of heat required The quantity of heat required

to raise the temperature of 1 to raise the temperature of 1 gram of a substance by one gram of a substance by one degree Celsius (or Kelvin)degree Celsius (or Kelvin)

The unit is J/g(The unit is J/g(ooC)C)

Some Specific Heat Some Specific Heat CapacitiesCapacities

SubstanceSubstance Specific heat Specific heat capacitycapacityin J/g(in J/g(ooC)C)

WaterWater 4.184.18CopperCopper 0.3850.385

IronIron 0.4490.449

Equation:Equation:

q = mC Tq = mC T q = heat transferred (in joules)q = heat transferred (in joules) m = mass (in grams)m = mass (in grams) C = specific heat capacityC = specific heat capacity T = change in temperatureT = change in temperature

T = TT = Tfinalfinal - T - Tinitialinitial

Practice ProblemPractice Problem

What is the What is the specific heatspecific heat of of silver if the silver if the temperature of temperature of a 15.4 ga 15.4 g sample of silver is sample of silver is increased from increased from 20.020.0ooCC to to 31.231.2ooCC when when 40.5 J40.5 J of heat of heat is added?is added?

Givens:Givens:m = 15.4 gm = 15.4 gTTii = 20.0 = 20.0ooCCTTff = 31.2 = 31.2ooCCq = 40.5 Jq = 40.5 J

q = mCq = mC∆T∆T40.5=15.4(C)(31.2-40.5=15.4(C)(31.2-

20.0)20.0)40.5=15.4(C)(11.2)40.5=15.4(C)(11.2)40.5=172.48(C)40.5=172.48(C) C = 0.235 J/g(C = 0.235 J/g(ooC)C)

Practice ProblemPractice ProblemWhat is the final temp of What is the final temp of silver if the temperature silver if the temperature of a 5.8 g sample of of a 5.8 g sample of silver starts out at 30.0silver starts out at 30.0ooC C and 40.5 J of heat is and 40.5 J of heat is added? The specific heat added? The specific heat of silver is .235 J/g(of silver is .235 J/g(ooC).C).

Givens:Givens:m = 5.8 gm = 5.8 gTTii = 30.0 = 30.0ooCCq = 40.5 Jq = 40.5 JC = 0.235 C = 0.235 TTff = ??? = ???

q = mCq = mC∆T∆T40.5=5.8(0.235)(T40.5=5.8(0.235)(Tf f -30.0)-30.0)40.5=1.363(T40.5=1.363(Tff-30.0)-30.0)40.5=1.363T40.5=1.363Tff – 40.89 – 40.8981.39=1.363Tf81.39=1.363Tf TTff = 59.7139 = 59.7139

TTff = 60. = 60.ooCC

What quantity of heat is What quantity of heat is required to raise the required to raise the temperature of 100 mL of temperature of 100 mL of water from 45.6water from 45.6C to 52.8C to 52.8C? C? The specific heat of water is The specific heat of water is 4.184 J/g(4.184 J/g(C) and water has C) and water has a density of 1.00 grams/mL.a density of 1.00 grams/mL.

Practice ProblemPractice Problem

Givens:Givens:q = ?q = ?V = 100. mLV = 100. mLTTii = 45.6 = 45.6ooCCTTff = 52.8 = 52.8ooCCC = 4.184C = 4.184ddwaterwater=1.00 =1.00 g/mLg/mL

100 mL = 100 g100 mL = 100 g

q = mCq = mC∆T∆Tq=100(4.184)(52.8-q=100(4.184)(52.8-

45.6)45.6)q=3012.48q=3012.48q=3010 Jq=3010 J

Heat of Fusion & Heat of Fusion & Heat of VaporizationHeat of Vaporization

Specific heat works great for the part Specific heat works great for the part of the heating and cooling process of the heating and cooling process where temperature is changing but where temperature is changing but what do we do to calculate Q when we what do we do to calculate Q when we are in the process of a phase change?are in the process of a phase change?

Answer: We use heat of fusion or heat Answer: We use heat of fusion or heat of vaporization!of vaporization!

Heat of VaporizationHeat of Vaporization The amount of energy needed to The amount of energy needed to

vaporize (l vaporize (l g) a specific amount g) a specific amount of a liquid at constant temperatureof a liquid at constant temperature

Unit is usually J/gUnit is usually J/g

Q = m(HQ = m(Hvv))

Example ProblemExample Problem A 115 g sample of liquid is boiled A 115 g sample of liquid is boiled

over a 10-minute period. As the over a 10-minute period. As the liquid boils, the temperature remains liquid boils, the temperature remains constant with 2.27 x 10constant with 2.27 x 1055 J of heat is J of heat is absorbed. At the end of the 10-absorbed. At the end of the 10-minute period, all of the liquid has minute period, all of the liquid has been boiled away. What was the been boiled away. What was the heat of vaporization of the liquid?heat of vaporization of the liquid?

GivensGivensQ = 2.27 x 10Q = 2.27 x 1055 J Jm = 115 gm = 115 g

Q = m(HQ = m(Hvv))2.27 x 102.27 x 1055 = 115(H = 115(Hvv))HHvv = 1973.91 J/g = = 1973.91 J/g = 1970 J/g1970 J/g

Heat of FusionHeat of FusionThe amount of heat needed The amount of heat needed

to change a specific amount to change a specific amount of a solid to a liquid at of a solid to a liquid at constant temperature.constant temperature.

Unit is usually J/gUnit is usually J/gQ = m(HQ = m(Hff))

Example ProblemExample ProblemThe heat of fusion of ice at The heat of fusion of ice at

00ooC is 3.4 x 10C is 3.4 x 1022 J/g. How J/g. How much heat is needed to much heat is needed to change 75g of ice at 0change 75g of ice at 0ooC to C to liquid water at the same liquid water at the same temperature?temperature?

GivensGivensHHff = 3.4 x 10 = 3.4 x 1022 J/g J/gm = 75 gm = 75 g

Q = m(HQ = m(Hff))Q = 75(3.4 x 10Q = 75(3.4 x 1022))Q = 2.55 x 10Q = 2.55 x 1044 J = 2.6 x 10 J = 2.6 x 1044 J J

QQTotalTotal

When using multiple equations to When using multiple equations to solve for the energy change of a solve for the energy change of a system, Qsystem, QTT is used. is used.

QQTT = Q = Q11 + Q + Q22 + etc. + etc.

Practice ProblemPractice Problem Water that has a mass of 283g is at Water that has a mass of 283g is at

23.523.5ooC and is placed in a freezer that C and is placed in a freezer that has a temperature of -12.4has a temperature of -12.4ooC. How C. How much energy must be removed from much energy must be removed from the water to reach this temperature?the water to reach this temperature?