Chapter 3.

Energy and the First Law

Soong Ho Um

Sungkyunkwan University

Chemical Engineering

Introduction

• Thermodynamic systems are characterized by an additional

state variable: temperature.

• Temperature is a measure of the energy stored inside matter

in various forms, which we collectively call internal energy. It

gives rise to another type of energy exchange that is not

encountered among purely mechanical systems, heat.

• The incorporation of heat effects into the energy balance

constitutes one of the fundamental principles of

thermodynamics known as the first law.

Instructional Objectives

• Formulate the mathematical statement of the first law for a closed system and learn how to:

1. Do energy balances in closed system.

2. Distinguish between path and state functions.

3. Use the steam tables to calculate internal energy and enthalpy.

4. Apply the energy balance to systems undergoing vaporization or condensation.

5. Perform calculations of internal energy and enthalpy in the ideal-gas state.

Energy and Mechanical Work

• Heat, work and energy are measured in the same units but

they represent different physical entities.

• Energy is the ability of a system to produce work, namely,

ability to cause the displacement of a force. It is a property of

the state, storable quantity.

• Example: an object with mass m resting on the floor. Potential

energy -> Kinetic energy; if no change on the location at

certain time, the energy is preserved until relocated.

• Work takes place when a force is displaced. It is exchanged

during a process and characterizes, not the state of the

system, but the transition of the system between states.

Energy -- Mechanical work -- Energy --

A system initially in equilibrium state A undergoes a process that brings it to

final state B; during this process the system exchanges work with the surroundings

so that the energy change of the system is equal to the amount of work exchanged:

∆EAB = W

Work represents energy that passes from one system into another.

Energy vs. Work

• Energy is storable; it remains in the system for as long as the state

of the system is preserved. It is a state function.

• Work is energy in transit; it appears when energy is passed from

one system into another.

• Work is not a storable quantity as such. Work that enters a system

must be stored in some form of energy.

• Work is associated with a direction from one system in which it

originates to another, where it is transferred to.

• Sign convention for work: work is positive if it enters the system,

negative if it exits.

Shaft work and PV work

• Shaft work ≈ mechanical work

• Another more subtle form of work is associated with

movement of the system boundaries, PV work. Can you

discriminate these (shaft work vs. PV work)?

Pex = P + δP (External pressure =

System pressure + increment; It

is a positive increment for

compression but it is a negative

increment for expansion.)

If the process is conducted in

a quasi-static manner, then

δP - 0 and Pex - P.

(Mechancially reversible process)

For a process that moves in

the direction of decreasing V

(compression), the area is

negative and the work is

positive.

Importantly, PV work

depends on the entire path

that connects the two states.

PV work is a path function

whose value depends not

only on the initial and final

states, but on the entire path. △?

Example 3.1: PV work in Expansion

A cylinder fitted with a piston contains 1 liters of gas. The

piston has a 1-in diameter and weighs 5 kg. How much

work is needed to expand the gas reversibly to twice its

volume against the pressure of the atmosphere (1.013

bar)?

Hint: the gas expands against the combined pressure of

the piston and the atmosphere. The weight of the piston is Mg.

The area of the piston is πD2/4.

The pressure(Pꞌ) exerted by the piston is Mg/(πD2/4).

The total pressure(P) in the cylinder is Pꞌ + P0, which remains constant during the process.

Example 3.2: PV work using the Soave-

Redlich-Kwong equation of state

Ethylene is compressed reversibly in a closed system. The

compression is conducted isothermally at 350 K, from initial

pressure 20 bar to final pressure 55 bar. Calculate the work

using the SRK equation of state.

Hint: estimate V points at the corresponding P points on the

PV diagram.

Internal Energy and Heat

• Molecules possess energy of various forms. It includes kinetic

energy due to the motion of the center of mass in space

(translational kinetic energy). In case of polyatomic molecules

possess rotational kinetic energy, vibrational energy, potential

energy (as a result of interaction between different molecules

as well as between atoms of the same molecule).

• Matter, regardless of chemical composition or phase, is

capable of storing energy internally. The combined storage

modes as internal energy (U) is a state function and for a pure

substance, it is a function of pressure and temperature.

• The molecular nature of matter gives rise to a different type of

energy transfer, heat.

Heat shares some important characteristics with work

1. It is a transient form of energy that is observed during a change of state

(process); once the system is in equilibrium with its surroundings there is

no net heat transfer because both system and surroundings are at the

same temperature.

2. As a transient form, heat is not a storable mode of energy: once it enters a

system, it is stored as internal energy. It is incorrect to say that a system

“contains heat,” or to speak of energy that “is converted into heat.”

3. It has a direction, from the system to the surroundings, or vice versa.

4. It is a path function whose value is determined by the entire path of the

process. This property of heat is not obvious at the moment but will

become so in the next section.

5. Heat is positive if it is transferred to the system from the surroundings.

First Law for a Closed System

• All material systems possess internal energy. They may

possess various other forms of energy as well. Changes

of a state invariably require the exchange of energy

between the system and the surroundings.

• The exchanges of energy in the state transition must be

in the form of work (PV, shaft, or both) and/or heat: ∆EAB

= Q + W. Energy balance should be expressed as, ∆E (U

+ Ek + Ep) = Q + W. The contribution of Ek and Ep is

negligible. Therefore, ∆U = Q + W (operator ∆ is a

shorthand notation for a change of a property between

two states).

Example 3.4: Analysis of Joule’s Experiment

In this example we consider a variation of Joule’s experiment: a

thermally insulated vessel contains 10 kg of water. The liquid is

stirred by an impeller driven by a 1 kW motor. If the motor runs

for 1 min and all of the work it produces is transferred to the

liquid, analyze the experiment of the basis of the first law and

report the relevant amounts of heat, work, and internal energy.

Ans (hint):

∆U = W

Constant-Volume Heating

• Let’s think of the process in which a closed system is

heated under constant volume.

• No PV work with the fixed volume of the system as well

as no shaft work.

• The first law gives, Q = ∆U (constant volume process).

• The amount of heat that is exchanged under constant

volume is equal to the change of internal energy,

provided that no shaft work is present.

Let’s consider elementary paths in different states

Example 3.5: Constant-volume cooling?

A sealed metal cylinder contains steam at 1 bar, 500 C. How

much heat must be removed at constant volume in order to

produce saturated vapor?

Ans (hint):

∆U = W

Steam table

Constant-Pressure Heating

Example 3.6: Constant-pressure cooling of steam?

8.5 kg of steam at 600 C, 15 bar, are cooled at constant

pressure by removing 6200 kj of heat. Determine the final

temperature and the amount of PV work that is exchanged with

the surroundings.

Ans (hint):

∆U = W

Steam table

Constant-Temperature Process

An isothermal process is one

conducted at constant

temperature. Experimentally,

it is conducted by placing the

system inside a heat bath.

Its path is represented by an

isotherm.

Example 3.7: Isothermal compression of steam?

Steam is compressed isothermally at 350 C in a closed system.

Compression is conducted reversibly from initial pressure 20 bar

to final pressure 40 bar. Determine the amounts of work and

heat exchanged between the system and its surroundings.

Ans (hint):

∆UAB = Q + W

Steam table

Sensible Heat-Heat Capacities

• In the absence of phase transitions, the exchange of heat is

accompanied with a temperature change. Such heat is called

sensible because it can be sensed with a thermometer.

• The amount of heat needed to produce a given temperature

change varies from one substance to another and is quantified by

the heat capacity.

• The amount of heat depends on the path of the heating process:

constant-volume and constant-pressure path.

Constant-Volume Heat Capacity

Even though the definition of the heat capacity was

motivated by heat, the heat capacity is a partial derivative

of a state function (internal energy), and itself a state

function.

Constant-Pressure Heat Capacity

Both Cp and Cv has dimensions of energy per mass per

temperature. The term specific heat is sometimes used

when the heat capacity is reported on a per-mass basis.

Why are Cv and Cp different?

• Mathematical view: they are different due to different

partial derivatives

• Thermodynamic view: they are different as associated

with heating along different paths

• Physical argument view: In constant-volume heating the

amount Qv is used to increase temperature by ∆T. In

constant-pressure heating the amount Qp is used to

increase temperature by the same amount and to

expand the volume against the surroundings.

Qp > Qv -- Cp > Cv

Example 3.8: Heat capacities and heat?

According to the relation, C = Q/T, heat capacities are defined in

terms of heat. Does this make Cv and Cp path functions?

Ans (hint):

No. Why?

Practical Utility of the Heat Capacities

1. They allow to calculate heat in constant-volume and

constant-pressure processes. It is useful in energy

balance.

2. They allow to calculate changes in internal energy and

enthalpy. It allow to calculate these properties using

equations rather than tables or to obtain their values in

states that are not found in tables.

Effect of Pressure and

Temperature on Heat Capacity

• As state functions, Cv and Cp depend on pressure and

temperature.

• Heat capacity is a strong function of pressure in the

vapor phase but almost independent of pressure in the

liquid.

• Cp is fairly sensitive on temperature in both liquid and

gas phase.

Ideal-Gas Heat Capacity

• Ideal-gas heat capacities have been compiled for a large

number of pure components and the data are usually

presented in the form of empirical correlations that give

Cpig as a function of temperature.

Heat of Vaporization

Lever rule for enthalpy:

H = xLHL + xVHV

xL + xV = 1

H = HV - xL∆Hvap

= HL + xV∆Hvap

Pitzer correlation for ∆Hvap

Example 3.9: Energy balances with phase change:

using tables

Steam at 1.013 bar, 200 C is cooled under constant pressure by

removing an amount of heat equal to 1000 kJ/kg. Determine the

final state.

Ans (hint):

H = Q

Ideal-Gas State

In the ideal-gas state, intermolecular interactions are unimportant

because distances between molecules are large, beyond the range

such interactions. If the volume of the system is increased at constant

temperature, there should be no change in internal energy: since

temperature remains constant, the kinetic energy of the molecules

(including translational, rotational, vibrational, etc.) is the same before

and after the expansion; and since there is no interaction between

molecules, there no other mode of energy storage available to the

system. Therefore, internal energy remains constant as long as

temperature is not changed. In other words, the internal energy in the

ideal-gas state is a function of temperature only:

Uig = Uig (T), Hig = Hig(T)

Reversible adiabatic process

Example 3.10: Heat Transfer

A 10 kg piece of hot copper at 450 C is quenched in an open tub

that contains 10 kg of water at 20 C. Calculate the final

temperature assuming no heat losses to the environment. The

heat capacities of copper and water are Cpc = 0.38 kJ/kg K, Cpw

= 4.184 kJ/ kg K, and may be assumed independent of

temperature.

Ans (hint):

Q = 0 = ∆H tot

Energy Balances and Irreversible Processes

• As a statement of energy conservation, the first law is

applicable to any process, whether it is reversible or not.

• If the process is irreversible and involves work, one must

be careful because dWrev = -PdV is not applicable.

• If the process does not involve work, the calculation is

done in the usual way.

• Implicit in all of these calculations is the assumption that

the system is internally uniform so that its state can be

described by a uniform pressure and temperature.

Example 3.11: Irreversible expansion against

vacuum

A closed insulated cylinder is divided into parts by a piston that is

held into place by latches. One compartment is evacuated, the

other contains steam at 7.5 bar, 300 C. The latches are removed

and the gas expands to fill the entire volume of the cylinder.

When the system reaches equilibrium, its pressure is 1 bar.

Determine the final temperature.

Ans (hint):

0 = ∆U12

Example 3.12: Irreversible expansion against

pressure

An insulated cylinder fitted with a weight less latched piston

contains steam at 5 bar, 300 C. The piston is unlatched and the

steam expands against ambient air at P =1 bar. The temperature

of steam in the final state is 142 C. Determine the energy

balance for this process.

Ans (hint):

W = ∆UAB

In this problem we were able to calculate the work from the

energy balance because ∆U and Q were both known.

Initial state Final state

P(bar) 5 1

T(C) 300 142

U(kJ / kg) 2803.3 2570.6