Section 4.4: Heat Capacity & Specific Heat

The Heat Capacity of a substance is defined as: C y (T)  (đQ/dT) y The subscript y indicates that property y of the substance is held constant when C y is measured.

Heat Capacity: C y (T)  (đQ/dT) y Specific Heat per Kilogram of mass m: mc y (T)  (đQ/dT) y Specific Heat per Mole of υ moles: υc y (T)  (đQ/dT) y

Heat Capacity SubstanceC Copper0.384 Wax0.80 Aluminum0.901 Wood2.01 Water4.18 Heat Capacity is different for each substance. It also depends on temperature, volume & other parameters. If substance A has a higher heat capacity than substance B, then More Heat is Needed to cause A to have a certain temperature rise ΔT than is needed to cause B to have the same rise in temperature.

Some Specific Heat Values

More Specific Heat Values

1 st Law of Thermo: đQ = dĒ + đW 2 nd 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: C y (T)  (đQ/dT) y gives the GENERAL RESULT: C y (T) = T(  S/  T) y

1 st 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: C V (T)  (đQ/dT) V = (  Ē/  T) V However, if the Pressure p is held constant, the 1 st Law must be used in the form đQ = dĒ + đW So, the Heat Capacity at Constant Pressure has the form: C p (T)  (đQ/dT) p

1 st Law of Thermodynamics: đQ = dĒ + đW = dĒ + pdV Heat Capacity at Constant Volume: C V (T)  (đQ/dT) V = (  Ē/  T) V Heat Capacity at Constant Pressure C p (T)  (đQ/dT) p NOTE!! Clearly, in general, C p ≠ C V Further, in general, C p > C V C p & C V are very similar for solids & liquids, but very different for gases, so be sure you know which one you’re using if you look one up in a table!

Heat Capacity Measurements for Constant Volume Processes (c v ) 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 1 st Law: Q = Ē 2 - Ē 1 =  Ē = mc v  T Heat Q added m m TT Insulation

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 = mc p  T Heat Q added TT m m xx Heat Capacity for Constant Pressure Processes (c p )

Experimental Heat Capacity Experimentally, it is generally easier to add heat at constant pressure than at constant volume. So, tables typically report C p for various materials.

Calorimetry Example: Similar to Reif, pages 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 Q s leaving the sample equals the heat energy that enters the water, Q w. This gives: Q s + Q w = 0 Typical Calorimeter

Q s + Q w = 0 (1) Sample Properties: Mass = m s. Initial Temperature = T s. Specific Heat = c s (unknown) Water Properties: Mass = m w. Initial Temperature = T w. Specific Heat = c w (4,286 J/(kg K)) Final Temperature (sample + water) = T f Assume c W & c s are independent of temperature T. Put Q s = m s c s (T f – T s ) & Q w = m w c w (T f – T w ) into (1): m s c s (T f – T s ) + m w c w (T f – T w ) = 0

Q s + Q w = 0 (1) m s c s (T f – T s ) + m w c w (T f – T w ) = 0 Solve for c s & get: l Technically, the mass of the container should be included, but if m w >> m container it can be neglected.

A slightly different problem. 2 substances A & B, initially at different temperatures T A & T B. Specific heats c A & c B are known. The final temperature T f is unknown. All steps are the same as the previous example until near the end! Q A + Q B = 0 (1) Sample Properties: Mass A = m A. Initial Temperature = T A. Specific Heat = c A Mass B = m B. Initial Temperature = T B. Specific Heat = c B Final Temperature (A + B) = T f (Unknown)

Q A + Q B = 0 (1) Assumptions: c A & c B are independent of temperature T. Put Q A = m A c A (T f – T A ) & Q B = m B c B (T f – T B ) into (1): m A c A (T f – T A ) + m B c B (T f – T B ) = 0 Solve for T f & get: T f = (m A c A T A + m B c B T B ) ∕(m A c A + m B c B )

Section 4.5: Entropy

The 2 nd 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: S b - S a = ∫dS = ∫(đQ/T) (1)

Entropy Difference Between Macrostate a & Macrostate b: ΔS = S b - S a = ∫dS = ∫(đQ/T) (1) Use the definition of heat capacity at constant y, C y (T)  (đQ/dT) y we can write đQ = C y (T)dT so that (1) becomes ΔS = S b - S a = ∫[(C y (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 = S b - S a = ∫[(C y (T)dT)/T] (2) ΔS is independent of the process by which the system goes from macrostate a to macrostate b, so this is very useful. For the very special case where the macrostate of the system is characterized only by it’s temperature T, with all other parameters y held fixed, (2) can be written: ΔS = S(T b ) – S(T a ) = ∫[(C y (T)dT)/T]

Sect. 4.6: Some Practical Consequences of the 3 rd 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 3 rd Law of Thermodynamics As the Temperature T  0, the Entropy S  S 0, S 0  A number independent of all system parameters. Often, S 0 = 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 ; 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 ) This problem is solved by using the result that, if the temperature dependence of the heat capacity C y (T) is known, the entropy difference at two temperatures can be calculated from ΔS = S(T b ) – S(T a ) = ∫[(C y (T)dT)/T] (1) & by using (1) along with the 3 rd Law of Thermodynamics!

Example 1 (pages ) ΔS = S(T b ) – S(T a ) = ∫[(C y (T)dT)/T] (1) Use (1) along with the 3 rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy S A  S A0 = S g0 & System B Entropy S B  S B0 = S w0, where S A0  S B0  S 0 or S g0  S w0  S 0 S w0  S g0  S 0 must be true, since The 3 rd Law says that S 0 is independent of the system parameters (so it is independent of the crystal phase of Sn!).

Example 2: 2 Systems: (pages ). 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 ) This problem is solved by using the result that, if the temperature dependence of the heat capacity C y (T) is known, the entropy difference at two temperatures can be calculated from ΔS = S(T b ) – S(T a ) = ∫[(C y (T)dT)/T] (1) & by using (1) along with the 3 rd Law of Thermodynamics

Example 2 (pages ) ΔS = S(T b ) – S(T a ) = ∫[(C y (T)dT)/T] (1) Use (1) along with the 3 rd Law of Thermodynamics That is, as the Temperature T  0, System A Entropy S A  S A0 = S Pb + S S & System B Entropy S B  S B0 = S PbS where S A0  S B0  S 0 or S Pb + S S  S PbS  S 0 S Pb + S S  S PbS  S 0 must be true, since The 3 rd Law says that S 0 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

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 y 1 & y 2, as in the Figure  : y1y1 y2y2 There are 2 possibilities: 1. If y = y 1 + y 2, the parameter y is an Extensive Parameter. 2. If y = y 1 = y 2, 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.