STATEMENTS OF THE SECOND LAW

B. STATEMENTS OF THE SECOND LAW

If a gas is adiabatically compressed from a state characterized by P 1 and v 1 to a state (P 2 ,v 2 ) through work transfer w 12 ( Figure 1 ), adiabatic expansion from state 3 (which is the same as state 2) to (P 4 ,v 4 ) (where state 4 is the same as state 1) provides work output |w 34 |= |w 12 |, i.e., w cyc = 0. If a higher pressure is required at the end of the expansion process then heat must be supplied at state 2 (e.g., at constant volume so that v 3 =v 2 and P 3 >P 2 ). The expansion to state 4 (for which v 4 =v 1 ) produces P 4 >P 1 . In this case the adiabatic curves 3-4 and 1-2 are almost parallel and, consequently, T 4 > T 1 . We make the point that heat rejection must close the cyclic process. The Otto cycle used in automobiles is schematically illustrated through part (d) of Figure 1 in the form of a P–v diagram. It consists of four processes: adiabatic quasie- quilibrium compression (1–2), constant volume heat addition (2–3), adiabatic quasiequilibrium expansion (3–4), and constant volume heat rejection (4–1).

The simplest statement of Second law is that for heat input a cyclic process requires heat rejection.

1. Informal Statements

We present two informal statements of the Second law. Statement 1: “The efficiency of a heat engine is less than unity.” The efficiency η of a thermodynamic cycle is defined as the ratio of the work output

to the thermal input, i.e., η = = Sought/Bought = W cycle /Q in .

(1) For a cyclical process, the First law states that ∫ δQ = Q in –Q out =W cycle = ∫ δW.

(2) Using Eqs. (1), (2) and the informal statement, η = (1– Q out /Q in ) < 1.

(3) Note that by using the subscripts “in” and “out”, the sign convention has already been

accounted for. For a finite work output, Eq. (3) implies that Q out is always nonzero. The ratio of the actual mechanical work to cycle work is called the system mechanical efficiency, and the product of the cycle and mechanical efficiencies is sometimes referred to as the thermal efficiency.

Statement 2: “An isolated system initially in a state of nonequilibrium will sponta- neously achieve an equilibrium state.”

A spontaneous process is one that occurs without outside intervention (i.e., without work or heat transfer). Consider a cup of warm water placed in a room made of rigid, insu- lated, and nonpermeable walls. Heat transfer occurs from the water and cup to room air until all three are at the same temperature. At this state equilibrium is reached. It is impossible to reheat the water and cup back to their initially higher temperature without external interven- tion, i.e., reheating is impossible by using the room air alone, since the final state consisting of the higher–temperature water (and cup) placed in lower–temperature air would be in a state of nonequilibrium, and in violation of the Second law. We will illustrate this concept using Ex- amples 1 and 2.

a. Example 1

A stirrer is used to warm 0.5 kg of water that is contained in an insulated vessel. The water is initially at a temperature of 40ºC. A pulley–weight assembly rotates the stir- rer through a gear mechanism such that a 100 kg mass falls through a height of 4220 cm. Is this process really possible? What is the final temperature? Assume that u = cT –1 –1

(with T expressed in ºC), and c = 4.184 kJ kg K . Solution The potential energy of the falling weight changes, and the consequent work input into the system heats the water. The potential energy change is

∆PE = 100×9.81×42.2÷1000 = 42 kJ.

The First law states that Q 12 –W 12 =U 2 –U 1 . At 40ºC,

U 1 = 0.5 ×40×4.184 = 84 kJ.

Since Q 12 = 0, ∆PE = –(–42) = U 2 –U 1 , or

U 2 = 42+84 = 126 kJ. Therefore, u 2 = 126÷0.5 = 252 kJ kg –1 , and T 2 = 252÷4.184 = 60ºC.

The process is possible and all the work is converted into “heat” (thermal energy)

Remarks Even though the internal energy cannot be measured directly, the pulley system en- ables us to implicitly calculate its values. Note that in this case all of the work can be converted into the thermal (heat) energy of water. Another example involves running a blender filled with ice cream over a long time period. Again, in this case, work input is converted into “heat” and the ice cream will melt. One can complete a cyclic process by rejecting heat to the ambient reservoir and cooling the water from 60 °C to 40 °C. In this cyclic process, the First law states that

∫ δQ - ∫δW = 0 or ∫δW =W cyc =∫ δQ < 0, i.e., a cyclic process can be completed with heat rejection to a single reservoir but with work input.

b. Example 2

42 kJ of heat energy are transferred from a thermal energy reservoir that exists at a temperature of 27ºC and are converted into work using a hypothetical heat engine un-

dergoing a cyclical process. The process raises a 100 kg weight through a 4220 cm height. The weight is then allowed to fall (as in Example 1) and the work is used to heat 0.5 kg of water that is initially at a temperature of 40ºC. Is such a scenario possi-

ble? What is the final water temperature? Assume u = cT, and c = 4.184 kJ kg –1 K –1 . Solution

In this scenario, the combined system that includes the heat engine and weight per- forms no net work, but can continuously transfer heat from the lower temperature thermal energy reservoir at 27ºC to warmer water that exists at 40ºC. This cannot be done through direct contact alone. Therefore, the combined system within the bound- ary E can be made to proceed further towards nonequilibrium, which is counter intui- tive and the process is impossible. However, this process can still be analyzed on the basis of the First law alone, i.e.,

– (– 42) = U 2W –U 1W , (A ) where, U 1W = 0.5 ×4.184 × 40 = 84 kJ. From Eq. (A),

U 2W = 42 + 84 = 126 kJ, u 2W = 252 kJ; T 2W = 60ºC. Remark

Recall that work can be converted into “heat” and hence the problem lies in the con- version of all of the “heat” into work.

a. Kelvin (1824-1907) – Planck (1858-1947) Statement

“It is impossible to devise a machine (i.e., a heat engine) which, operating in a cy- cle, produces no effect other than the extraction of heat from a thermal energy res- ervoir and the performance of an equal amount of work.”

The First law conserves energy, and the Second law prohibits the complete conver- sion of thermal energy into work during a cyclical process. Using the First law for a closed system undergoing a cyclical process in a heat engine, ( ∫ δQ = ∫ δW) > 0. Note that the Kel-

vin-Planck statement does not preclude the condition ∫ δW ≈ 0 (cf. Example 1). Consider an adiabatic mass of water being heated by the action of a frictionless stirrer that raises the water

temperature. If the insulation and stirrer are removed, it is possible to cool the water to its ini- temperature. If the insulation and stirrer are removed, it is possible to cool the water to its ini-

Therefore, heat energy (i.e., Q = U 2 –U 1 ) possesses a lower quality than an equal amount of work energy, since it is capable of a smaller amount of useful work.

b. Clausius (1822-1888) Statement Heat cannot flow from a lower to higher temperature. However, heat can be transferred from a lower-temperature thermal energy reservoir to a higher-temperature reservoir in the presence of work input. The Clausius statement (due to Rudolf Clausius, 1822–1888) regarding this is as follows:

“It is impossible to construct a device that operates in a thermodynamic cycle and produces no effect other than the transfer of heat from a cooler to a hotter body.” An air conditioner transfers heat from the lower temperature indoor space of a house to a higher–temperature ambient during the summer. A heat pump delivers heat from a lower–temperature ambient to a higher–temperature system (such as a house). For example, using a heat pump, if 150 kJ of heat is transferred from cold air at, say, 0ºC in conjunction with

a work input of 100 kJ, the pump is capable of delivering 250 kJ of heat to a space. Rather than the efficiency, a refrigeration cycle is characterized by a Coefficient of Performance, namely

COP refrigeration = Energy Sought/Energy Bought = Heat transferred from the lower temperature system/Work input.

(4) In the case of a heat pump, COP heat pump = Energy Sought/Energy Bought

= Heat transferred to the higher temperature system/Work input. (5) Heat pumped at the rate of 3.516 kW (200 BTU/min or 211 kJ/mim) from a system

constitutes a unit that is referred to as one ton of capacity. The physical implication is derived from the cooling of water, i.e., if 3.516 kW of thermal energy is removed from 1 ton of liquid water at 0ºC, transforming it completely into ice at 0ºC over a duration of 24 hours. Instead of (COP) cooling which is dimensionless, industries use a unit called HP/ton of refrigeration (1 HP

= 550 ft lb f /s = 0.7457 kW= 42.42 BTU/min, HP/Ton = 4.715/COP). Employing Eqs. (2), (4), and (5)

COP refrigeration =Q in from lower T system ÷|Q in from lower T system –Q out to higher T system |

(7) Consider a system undergoing a cyclical process and pumping heat from a lower-

COP heat pump =Q out to higher T system ÷|(Q in from lower T system –Q out to higher T system |

temperature thermal energy reservoir to a higher-temperature thermal energy reservoir. From the First law ( ∫ δQ = ∫ δW) < 0, since there is work input into such a process. Therefore, for

a refrigeration cycle, such as one for an air conditioner (– Q H +Q L ) < 0, where Q H denotes the heat leaving a system and entering a warmer thermal energy reservoir (e.g., the ambient), and Q L is that entering the system from a cooler thermal energy reservoir (e.g., a cooled residential space).

i. Perpetual Motion Machines

A machine that obeys the First law but violates the Second law of thermodynamics is known as a perpetual motion machine of the second kind (PMM2). Other terms for it include anti–Clausius machine and anti–Kelvin machine.

Hot Pressure Cooker at 500 K

HP-B

300 K kJ

kJ 150

Ambience

kJ kJ HE-A

kJ

50 kJ

Figure 2: An example of a PMM2.

c. Example 3 Consider the following hypothetical scenario based solely on the First law. An insu- lated pressure vessel contains superheated steam. The thermal energy contained in the steam is converted into work in order to run a heat engine A (Q in,A ,Q out,A ,W cycle,A ).

However, this decreases the steam temperature, and a heat pump B (Q in,B ,Q out,B ,W cy- cle,B ) is employed to pump thermal energy into the vessel to raise the temperature to its initial value, such that Q out,B =Q in,A . Note that the numbers carry a positive sign for all the symbols. Now, W cycle,A =Q in,A –Q out,A , and W cycle,B =Q out,B –Q in,B =Q in,A – Q in,B . Therefore, W wheels =W cycle,A –W cycle,B =Q in,B –Q out,A. If Q in,B >Q out,A , we can harness the difference W cycle,A –W cycle,B in order to run an automobile. Is it possible to operate an automobile perpetually in this manner without consuming fuel and,

thereby, emitting no pollutants or greenhouse gases? (See Figure 2 .)

Solution Consider the case Q out,A = 0 . This is a violation of the Second law. A consequence is that with Q in,A = 150 kJ, W cycle,A = 150 kJ, Q in,B = 50 kJ, Q out,B =Q in,A = 150 kJ, W cy- cle,B = 100 kJ, W cycle,A -W cycle,B =Q in,B –Q out,A = 50 kJ. The question is whether one can use the energy from the ambient to run an automobile. (Note that frictional work transferred at the tires can be used to supply energy back into the ambient.) This is an example of a perpetual motion machine of the second kind. The heat engine A vio- lates the Kelvin–Planck statement of the Second law, and its existence is not possible. If Q out,A = 45 kJ, the machine A is not in violation of the Second law. In this case W wheels =Q in,B –Q out,A = 5 kJ. However, for the dashed boundary, 5 kJ of energy en- ters from the ambient and it is all converted into work. Again, this violates the Kel- vin–Planck statement of the Second law. Thus, the system within the dashed boundary in Figure 2 remains unchanged during both of these processes, and the entire heat crossing the boundary is converted into work. This is impossible.