Section 4.4: Heat Capacity & Specific Heat

•The Heat Capacity of asubstance is defined as:

Cy(T) (đQ/dT)y

•The subscript y indicates thatproperty y of the substance isheld constant when Cy is

measured.

Heat Capacity: Cy(T) (đQ/dT)y

•Specific Heat per Kilogramof mass m:

mcy(T) (đQ/dT)y

•Specific Heat per Moleof υ moles:

υcy(T) (đQ/dT)y

Heat CapacitySubstance C

Copper 0.384Wax 0.80

Aluminum 0.901Wood 2.01Water 4.18

•Heat Capacity is differentfor each substance. It alsodepends on temperature,volume & other parameters.

•If substance A has a higher heat capacitythan substance B, then More Heat isNeeded to cause A to have a certaintemperature rise ΔT than is needed to causeB to have the same rise in temperature.

Some Specific Heat Values

More Specific Heat Values

More Specific Heat Values

1st Law of Thermo: đQ = dĒ + đW 2nd Law of Thermo: đQ = TdS

dS = Entropy Change• Combining these gives:

TdS = dĒ + đW• Using this result with the definition of Heat

Capacity at constant parameter y:

Cy(T) (đQ/dT)y

gives the GENERAL RESULT:

Cy(T) = T(S/T)y

1st Law of Thermo: đQ = dĒ + đW• If volume V is the only external parameter,

đW = pdV. •Under constant volume conditions (dV = 0)

đQ = dĒ •So, the Heat Capacity at Constant Volume has

the form: CV(T) (đQ/dT)V = (Ē/T)V

•However, if the Pressure p is held constant, the1st Law must be used in the form đQ = dĒ + đWSo, the Heat Capacity at Constant Pressure

has the form: Cp(T) (đQ/dT)p

•1st Law of Thermodynamics: đQ = dĒ + đW = dĒ + pdV

•Heat Capacity at Constant Volume:CV(T) (đQ/dT)V = (Ē/T)V

•Heat Capacity at Constant PressureCp(T) (đQ/dT)p

NOTE!! Clearly, in general, Cp ≠ CV

Further, in general, Cp > CV

•Cp & CV are very similar for solids & liquids, but

very different for gases, so be sure you knowwhich one you’re using if you look one up in a table!

Heat Capacity Measurements for Constant Volume Processes (cv)

• Heat is added to a substance of mass m in a fixed volume enclosure, which causes a change in internal energy, Ē. So, from The 1st Law:

Q = Ē2 - Ē1 = Ē = mcvT

Heat Qaddedm m

TInsulation

• Heat is added to a substance of mass m held at a fixed pressure, which causes a change in internal energy, Ē, AND some work pV.

Q = Ē + W = mcpT

Heat Qadded

T

m m

x

Heat Capacity for Constant Pressure Processes (cp)

Experimental Heat Capacity

Experimentally, it is generally easier to add heat at constant pressure than at constant volume. So, tables typically report Cp for various materials.

Calorimetry Example: Similar to Reif, pages 141-142

• A technique to Measure Specific Heat is to heat a sample of material, add it to water, & record the final temperature. This technique is called

Calorimetry.• Calorimeter: A device in which

heat transfer takes place.• Conservation of Energy requires

that the heat energy Qs leaving the sample equals the heat energy that enters the water, Qw. This gives:

Qs + Qw = 0

Typical Calorimeter

Qs + Qw = 0 (1)• Sample Properties:

Mass = ms. Initial Temperature = Ts.

Specific Heat = cs (unknown)• Water Properties:

Mass = mw. Initial Temperature = Tw.

Specific Heat = cw (4,286 J/(kg K))• Final Temperature (sample + water) = Tf

• Assume cW & cs are independent of temperature T.• Put Qs = mscs(Tf – Ts ) & Qw = mwcw(Tf – Tw) into (1):

mscs(Tf – Ts ) + mwcw(Tf – Tw) = 0

Qs + Qw = 0 (1)mscs(Tf – Ts ) + mwcw(Tf – Tw) = 0

• Solve for cs & get:

ll

• Technically, the mass of the container should be included, but if mw >> mcontainer it can be neglected.

w w f ws

s s f

m c T Tc

m T T

• A slightly different problem. 2 substances A & B, initially at different temperatures TA & TB. Specific heats cA & cB are known. The final temperature Tf is unknown. All steps are the same as the previous example until near the end!

QA + QB = 0 (1)• Sample Properties:

Mass A = mA. Initial Temperature = TA.Specific Heat = cA

Mass B = mB. Initial Temperature = TB. Specific Heat = cB

• Final Temperature (A + B) = Tf (Unknown)

QA + QB = 0 (1)

• Assumptions: cA & cB are independent of temperature T.

• Put QA = mAcA(Tf – TA) &

QB = mBcB(Tf – TB) into (1):

mAcA(Tf – TA ) + mBcB(Tf – TB) = 0

• Solve for Tf & get: Tf = (mAcATA + mBcBTB) ∕(mAcA + mBcB)

Section 4.5: Entropy

The 2nd Law of Thermodynamics:(Infinitesimal, quasi-static process)

đQ = TdS or dS = (đQ/T)•If the system is brought quasistatically

between macrostate a & macrostate b,

The Entropy Difference is:

Sb - Sa = ∫dS = ∫(đQ/T) (1)

Entropy Difference Between Macrostate a & Macrostate b:

ΔS = Sb - Sa = ∫dS = ∫(đQ/T) (1)•Use the definition of heat capacity at constant y,

Cy(T) (đQ/dT)y

we can write đQ = Cy(T)dT so that (1) becomes

ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2) •(2) is very useful because ΔS is independent of the process by which the system goes from macrostate a to macrostate b.

•For a system which goes quasi-statically from macrostate a to macrostate b, the entropy change is:

ΔS = Sb - Sa = ∫[(Cy(T)dT)/T] (2)•ΔS is independent of the process by whichthe system goes from macrostate a tomacrostate b, so this is very useful. •For the very special case where the macrostateof the system is characterized only by it’s

temperature T, with all other parameters y held fixed, (2) can be written:

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T]

Sect. 4.6: Some Practical Consequences of the 3rd Law of

Thermodynamics

• As we’ve seen, for all applications, only differences in the internal energy ΔĒ matter. This is because, as we know, The Zero of Energy is Always Arbitrary.

• Similarly, for many applications, only Entropy DIFFERENCES ΔS are important. However, unlike internal energy, The ABSOLUTE Entropy S of a system can be known in principle from the

3rd Law of Thermodynamics• As the Temperature T 0, the Entropy S S0,

S0 A number independent of all system parameters. Often, S0 = 0

• This fact can sometimes be put to practical use, as the following two examples (Reif, pages 145 – 148) illustrate.

Example 1: 2 Systems:(pages 145-147; overview only!)

• System A: 1 Mole of Solid Gray Tin (α-Sn)(at room temperature & pressure)

• System B: 1 Mole of Solid White Tin (β-Sn)(at room temperature & pressure)

• The physical process to convert System A into System B is HORRENDOUSLY COMPLICATED!!

• However in this example,

The Entropy Difference at Room Temperature & Pressure Between System

A & System B can be Calculated!

Example 1 (pages 145-147)• This problem is solved by using the result

that, if the temperature dependence of the heat capacity Cy(T) is known, the entropy difference at two temperatures can be calculated from

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1)

& by using (1) along with the

3rd Law of Thermodynamics!

Example 1 (pages 145-147)

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1)• Use (1) along with the

3rd Law of Thermodynamics• That is, as the Temperature T 0,

System A Entropy SA SA0 = Sg0 &

System B Entropy SB SB0 = Sw0, where SA0 SB0 S0 or Sg0 Sw0 S0

• Sw0 Sg0 S0 must be true, since The 3rd Law says that S0 is independent of the system parameters

(so it is independent of the crystal phase of Sn!).

Example 2: 2 Systems:(pages 147-148).

• System A: 1 Mole of Solid Lead (Pb) + 1 Mole of Solid Sulfur (S)

(at room temperature & pressure)

• System B: 1 Mole of Solid Lead Sulfide (PbS). • The chemical reaction to convert System A into

System B is HORRENDOUSLY COMPLICATED!!• However in this example,

The Entropy Difference at Room Temperature Between System A& System B can be Calculated!

Example 2 (pages 147-148)

• This problem is solved by using the result that, if the temperature dependence of the heat capacity Cy(T) is known, the entropy difference at two temperatures can be calculated from

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1)

& by using (1) along with the

3rd Law of Thermodynamics

Example 2 (pages 147-148)

ΔS = S(Tb) – S(Ta) = ∫[(Cy(T)dT)/T] (1)• Use (1) along with the

3rd Law of Thermodynamics• That is, as the Temperature T 0,

System A Entropy SA SA0 = SPb + SS & System B Entropy SB SB0 = SPbS

where SA0 SB0 S0 or SPb + SS SPbS S0

• SPb + SS SPbS S0 must be true, since The 3rd Law says that S0 is independent of the system

parameters (so it is independent of the chemical environment of the Pb & S atoms!)

Section 4.7:Extensive & Intensive Parameters

Extensive & Intensive Parameters• There are two types of macroscopic parameters which can

specify the system macrostate. Lets illustrate as them as follows.• Let y be such a parameter. Start with a single system for

which y has a certain value, as in the Figure y

• Now, divide the system into two parts with a partition.• The values of the parameter y for the two parts are y1 &

y2, as in the Figure : y1 y2

There are 2 possibilities:1. If y = y1 + y2, the parameter y is an Extensive

Parameter.2. If y = y1 = y2, the parameter y is an Intensive

Parameter.

• In other words:

An Extensive Parameter is:• A system parameter that depends on the

extent or size of the system.Examples: Mass m, Volume V, Internal Energy Ē,

Heat Capacity C, Entropy S, ...

An Intensive Parameter is:• A system parameter that independent of the

extent or size of the system. Examples: Temperature T, Mass Density ρ,

Specific Heat c, Pressure p, ...

The ratio of any 2 extensive parameters is intensive.