PHY203: Thermal Physics Topic 4: Second Law of Thermodynamics Heat Engines Statements of the Second Law (Kelvin, Clausius) Carnot Cycle Efficiency of a Carnot engine Carnot’s theorem Introduction to idea of entropy Entropy as a function of state Entropy form of second law Example calculations

Heat Engines Heat engines operate in a cycle, converting heat to work then returning to original state at end of cycle. A gun (for example) converts heat to work but isn’t a heat engine because it doesn’t operate in a cycle. In each cycle the engine takes in heat Q1 from a “hot reservoir”, converts some of it into work W, then dumps the remaining heat (Q2) into a “cold reservoir” HOT COLD Engine Q1 Q2 W

HOT COLD Engine Q1 Q2 W Efficiency of a heat engine Definition: Because engine returns to original state at the end of each cycle,  U(cycle) = 0, so W = Q1 - Q2 Thus:

Efficiency of a heat engine According to the first law of thermodynamics (energy conservation) you can (in principle) make a 100% efficient heat engine. BUT…………. The second law of thermodynamics says you can’t:

Kelvin Statement of Second Law: “No process is possible whose SOLE RESULT is the complete conversion of heat into work” William Thomson, Lord Kelvin ( )

HOTCOLD HOTTERCOLDERWARM Q HOTCOLD Q Both processes opposite are perfectly OK according to First Law (energy conservation) But we know only one of them would really happen – Second Law Heat flow

Clausius Statement of Second Law: “No process is possible whose SOLE RESULT is the net transfer of heat from an object at temperature T 1 to another object at temperature T 2, if T 2 > T 1 ” Rudolf Clausius ( )

How to design a “perfect” heat engine 1)Don’t waste any work So make sure engine operates reversibly (always equilibrium conditions, and no friction). 2)Don’t waste any heat So make sure no heat is used up changing the temperature of the engine or working substance, ie ensure heat input/output takes place isothermally Sadi Carnot ( )

Hot Source T 1 Q 1 Piston Gas a b Working substance (gas) expands isothermally at temperature T 1, absorbing heat Q 1 from hot source. The Carnot Cycle (I): isothermal expansion T1T1

The Carnot Cycle (II): adiabatic expansion Gas isolated from hot source, expands adiabatically and temperature falls from T 1 to T 2. Piston Gas b c Gas isolated from hot source, expands adiabatically, and temperature falls from T 1 to T 2

The Carnot Cycle (III): isothermal compression 3) Gas is compressed isothermally at temperature T 2 expelling heat Q 2 to cold sink. Piston Gas d c Q 2 Cold Sink T 2 Gas is compressed isothermally at temperature T 2 expelling heat Q 2 to cold sink. T2T2

The Carnot Cycle (IV): adiabatic compression Gas is compressed adiabatically, temperature rises from T 2 to T 1 and the piston is returned to its original position. The work done per cycle is the shaded area. Piston Gas a d Gas is compressed adiabatically, temperature rises from T 2 to T 1 and the piston is returned to its original position. Work done is the shaded area.

Efficiency of ideal gas Carnot engine We can calculate the efficiency using our knowledge of the properties of ideal gases

Isothermal expansion (ideal gas)

Isothermal compression (ideal gas)

Adiabatic processes Efficiency of ideal gas Carnot engine

Can you do better than a Carnot? HOT COLD Carnot Q1 Q2 W “Super Carnot” Q3 Q4 W

HOT COLD Engine Q1 Q2 W HOT COLD Engine Q1 Q2 W A Carnot engine is reversible…….. …..so you can drive it backwards

HOT COLD Carnot Q1 Q2 W “Super Carnot” Q3 Q4 Drive Carnot backwards with work output from “Super Carnot” Heat leaving hot reservoir =Q3 – Q1, which is negative So, net heat enters the hot reservoir Since the composite engine is an isolated system, this heat can only have come from the cold reservoir Net result, transfer of heat from cold body to hot body, FORBIDDEN BY CLAUSIUS STATEMENT OF SECOND LAW

HOT COLD Carnot Q1 Q2 W “Super Carnot” Q3 Q4 Carnot’s Theorem: “No heat engine operating between a hot (T h ) and a cold (T c ) reservoir can be more efficient than a Carnot engine operating between reservoirs at the same temperatures” It follows, by exactly the same argument, that ALL Carnot engines operating between reservoirs at same T c, T h, are equally efficient (ie independent of “working substance”) – our ideal gas result holds for all Carnot Cycles

So…….. Conservation of “Q/T” For all Carnot Cycles, the following results hold: What about more general cases?????

Was derived from expressions for efficiency, where only the magnitude of the heat input/output matters. If we now adopt the convention that heat input is positive, and heat output is negative we have: The expression

Arbitrary reversible cycle can be built up from tiny Carnot cycles (CCs) Q1Q1 Q3Q3 Q2Q2 Q4Q4 For 2 CCs shown: For whole reversible cycle: In infinitesimal limit,  Q → dQ,  → ,fit to cycle becomes exact: For any reversible cycle: Arbitrary reversible cycle

Entropy To emphasise the fact that the relationship we have just derived is true for reversible processes only, we write: We now introduce a new quantity, called ENTROPY (S) Entropy is conserved for a reversible cycle

Is entropy a function of state? P V A B path1 path2 For whole cycle: Entropy change is path independent → entropy is a thermodynamic function of state Reversible cycle

Example calculation: Calculate the entropy change of a 10g ice cube at an initial temperature of -10°C, when it is reversibly heated to completely form liquid water at 0°C…….. Specific heat capacity of ice = 2090 J kg -1 K -1 Specific latent heat of fusion for water = 3.3  10 5 J kg -1

Irreversible processes Carnot EngineIrreversible Engine For irreversible case:

Irreversible processes Following similar argument to that for arbitrary reversible cycle: P V A B Path 1 (irreversible) Path 2 (reversible ) For irreversible cycle Irreversible cycle

Irreversible processes Equality holds for reversible change, inequality holds for irreversible change General Case

“Entropy statement” of Second Law We have shown that: For a thermally isolated (or completely isolated) system, dQ = 0 “The entropy of an isolated system cannot decrease”

What is an “isolated system” The Universe itself is the ultimate “isolated system”, so you sometimes see the second law written: “The entropy of the Universe cannot decrease” (but it can, in principle, stay the same (for a reversible process)) It’s usually a sufficiently good enough approximation to assume that a given system, together with its immediate surroundings constitute our “isolated system” (or universe)………

Entropy changes: a summary For a reversible cycle:  S(system) =  S(surroundings) = 0  S(universe) =  S(system) +  S(surroundings) = 0 For a reversible change of state (A → B ) :  S(system) = -  S(surroundings) = not necessarily 0  S(universe) =  S(system) +  S(surroundings) = 0 For an irreversible cycle  S(system) = 0;  S(surroundings) > 0 For a irreversible change of state (A → B ) :  S(system) ≠ -  S(surroundings)  S(universe) =  S(system) +  S(surroundings) > 0

Example: entropy changes in a Carnot Cycle

Entropy calculations for irreversible processes Suppose we add or remove heat in an irreversible way to our system, changing state from A to B At first sight we might think it’s a problem to use the relation: However, because S is a function of state,  S for the change of state must be the same for all paths, reversible or not. So, we can “pretend” the heat required to produce the given change of state was added or removed reversibly and use the formula anyway! BUT… the entropy change of the surroundings must be different for the reversible (  S(universe)=0) and irreversible (  S(universe)>0) cases.

Entropy calculations for adiabatic processes Here, we have no “dQ” term at all, so where do we start with the calculation……….. To calculate the entropy change of the system we can “invent” a non- adiabatic process that takes the system between the same 2 states as our actual, adiabatic process and use that “dQ” to do the calculation. Again we rely on the path independence of  S Example: Joule expansion of an ideal gas: (irreversible, adiabatic change of state) Before After Rigid, adiabatic wall GAS V i GAS V f

Before After Rigid, adiabatic wall GAS V i GAS V f Joule expansion of an ideal gas For this process,  U(gas)=0 For ideal gas, U is a function of T only, so  T = 0 So, our “model” process is a reversible, isothermal expansion from V i to V f (  S(gas) = nRln(V f /V i ), see Carnot cycle calculation) But, for Joule expansion,  S(universe) =  S(gas): gas is an isolated system For reversible isothermal expansion,  S(surroundings) =-nRln(V f /V i ),  S(universe) = 0

Some past (part) exam questions…. 1)A copper block of heat capacity 150J/K, at an initial temperature of 60°C is placed in a lake at a temperature of 10°C. Calculate the entropy change of the universe as a result of this process 2)A thermally insulated 20  resistor carries a current of 10A for 1 second. The initial temperature of the resistor is 10  C, its mass is 5x10 -3 kg, and its heat capacity is 850 Jkg -1 K -1. Calculate the entropy change in a) the resistor and b) the universe.