Thermodynamics

Energy is...• The ability to do work.• Conserved.• Made of heat and work.• A state function: values that depend on the state of the

substance, and not on how that state was reached. • Independent of the path, or how you get from point A to B.• Work is a force acting over a distance.• Heat is energy transferred between objects because of

temperature difference.

The Universe

• Is divided into two halves.• The system and the surroundings.• The system is the part you are concerned with.• The surroundings are the rest.• Exothermic processes release energy to the surroundings.• Endothermic processes absorb energy from the

surroundings.

Units of Heat

• Calorie (cal)– The quantity of heat required to change the

temperature of one gram of water by one degree Celsius.

• Joule (J)– SI unit for heat

1 cal = 4.184 J

Direction

Every energy measurement has three parts.1. A unit ( Joules or calories).2. A number how many (magnitude).3. A sign to tell direction.• negative - exothermic• positive- endothermic

System

Surroundings

Energy

DE <0

System

Surroundings

Energy

DE >0

First Law of Thermodynamics

• The energy of the universe is constant.• Law of conservation of energy.• q = heat• Take the system’s point of view to decide signs.• q is negative when the system loses heat energy• q is positive when the system absorbs heat energy

Heat Capacity• The quantity of heat required to change the temperature

of a system by one degree.

• Three “types” with different systems:– Specific Heat Capacity– Molar Heat Capacity (we won’t cover this one in Physics)– Heat Capacity

Heat Capacity Types

– Specific heat capacity• System is one gram of substance• Units are J/goC• q = heat change• m = mass• C = Specific Heat Capacity• = DT final temperature – initial temperature

q = (m)(C)(T)

Heat Capacity Types

– Heat capacity of an object• Specific for a given amount or unit

so it is not mass dependent• Units are J/oC• q = heat• C = heat capacity• = DT final temperature – initial temperature

q = CT

Heat Capacity Types

• So how do you know which one to use when?• Look at the known information• Select the value with appropriate units to

cancel

Example

• The specific heat of graphite is 0.71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

• The specific heat of graphite is 0.71 J/goC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

q = mCΔT = (75,000 g)(0.71 J/goC)(348-294)K= 2.88 x 106 J

Note that ΔT oC and ΔT K are interchangeable, since they are the same “size”.

Example

Example

When a piece of copper (5.0 g) is heated for 2.0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20.0 ˚C to 71.9 ˚C. What is the specific heat of the copper?

m = 5.0 g; ΔT = (71.9 – 20.0) = 51.9˚C; q = 100 J; t = 2.0 sq = mCΔT; q = C = 100 = 0.385 J/goC

mΔT (5.0)(51.9)

time is irrelevant.

Example

If 10.0 g of Cu is heated for 2.0 seconds from 20.0 ˚C and 200 J of heat are absorbed, what is the final temperature of the block?(Let’s assume time is again irrelevant.)m = 10.0 g; Ti = 20.0 oC; Tf = ?; CCu = 0.385 J/goC; q = 200 J

q = mCΔT; ΔT = q/mC = (Tf – Ti); Tf = q/mC + Ti

Tf = [200/(10*0.385)] + 20 = 71.9oC

Calorimetry

• Measuring temperature changes to calculate heat changes.

• Use a calorimeter.• Two kinds– Constant pressure calorimeter (called a coffee cup

calorimeter)– Constant volume calorimeter (called a bomb

calorimeter)

Calorimetry

• A coffee cup calorimeter measures temperature and calculates q.

• An insulated cup, full of water (constant pressure, variable volume)

• Water is the surroundings in which the system changes• Calculate the heat change of water. • The specific heat of water is 4.184 J /goC• Heat of water qH2O = CH2O mH2O DT

qsystem + qsurroundings = 0

• qH2O + qrxn = 0

• qH2O = ‒ qrxn

In interactions between a system and its surroundings the total energy remains constant— energy is neither created nor destroyed.

Law of Conservation of Energy

Coffee Cup Calorimeter

• A simple calorimeter.– Well insulated and therefore isolated.– Measure temperature change.

qrxn = ‒ qcal

Determination of Specific Heat

ExampleDetermining Specific Heat from Experimental Data.

Use the data presented on the last slide to calculate the specific heat of lead.

q Pb = ‒ q H2O

q H2O = mcT = (50.0 g)(4.184 J/g °C)(28.8 ‒ 22.0)°C

q H2O = 1.4x103 J

q Pb = ‒1.4x103 J = mcT = (150.0 g)(c)(28.8 ‒ 100.0)°C

c Pb = 0.13 Jg‒1°C‒1

ExampleEqual masses of liquid A, initially at 100 ˚C, and liquid B, initially at 50 ˚C, are combined in an insulated container. The final temperature of the mixture is 80 ˚C. Which has the larger specific heat capacity, A or B?

mA = mB = m; TiA = 100˚C; TiB = 50˚C; Tf = 80˚C

qA = qB; mCAΔTA = mCBΔTB; CAΔTA = CBΔTB; CA/CB = ΔTB/ΔTA

CA/CB = (80–50)/(100–80) = 1.5. Since CA/CB > 1, CA > CB

You do one…When 86.7 grams of water at a temperature of 73.0 ˚C is mixed with an unknown mass of water at a temperature of 22.3 ˚C the final temperature of the resulting mixture is 61.7 ˚C. What was the mass of the second sample of water? a. 24.9 g c. 48.2 g

b. 302 g d. 419 g

States of Matter

General Heating Curve

Changes of State for Water

Heat of vaporization (liquid <----> gas)

H2O(l) ↔ H2O(g)

Hv = 2,258 J/g

Heat of fusion (solid <----> liquid )

H2O(s) ↔ H2O(l)

Hf = 333 J/g

Heating Curve for WaterTotal heat absorbed by the water as it is warmed can be determined by finding the heat involved in each “step” of the process.

q1+ q2 + q3 …. = qtotal

Note:

Cice= Csteam = 2.09 J/goC

q = mCTqphase change = Hphase change

Example

Calculate q for the process in which 50.0 g of water is converted from liquid at 10.0°C to vapor at 125.0°C.

Example

• What is the heat of fusion of lead in J/g if 6.30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a liquid? a. 0.0250 J/g c. 1.61 J/g

b. 24.7 J/g d. 40.5 J/g

ExampleWhat quantity of heat is required to heat 1.00 g of lead from 25 ˚C to the melting point (327 ˚C) and melt all of it? (The specific heat capacity of lead is 0.159 J/g • K and it requires 24.7 J/g to convert lead from the solid to the liquid state.) a. 2.47 J c. 39.4 J

b. 48.0 J d. 72.7 J