Chapter 18

Heat and the First Law of Thermodynamics

Heat is the transfer of energy due to the difference in temperature. The

internal energy is the total energy of the object in its center-of-mass reference

frame. The amount of heat Q necessary to increase the temperature is

proportional to the temperature difference and the mass of the object,

,

where C is the heat capacity and c = C/m is called the specific heat.

1 calorie is the amount of heat needed to increase the temperature of 1 g of

water by 10 C, 1 cal = 4.184 J.

In the US heat is often measured in Btu (British thermal unit – amount of heat

needed to increase the temperature of 1 pound of water by 10 F,

� = Δ���� = ΔT = mc ΔT

1 Btu = 252 cal = 1.054 kJ

The specific heat of water1

2

��� � = 1 ��� × � = 4.184 ��

�� × �

The heat capacity per mole is called the molar heat capacity, � = /� = mc/n

Example 1. The specific heat of bismuth is 0.123 kJ/kg×K. How much heat is

needed to raise the temperature of 2kg of bismuth from 200C to 5000C?

� = mc ΔT = 2kg × 0.123 kJ/kg×K × 480 K = 118.08 kJ

Calorimetry is measuring specific heat using calorimeters (isolated thermal baths

of known heat capacity).

Example 2. One puts 1.2kg of lead initially at 100.70C into aluminum calorimeter

(mass 400g) with 1kg of water initially at 180C. What is the specific heat of lead

if the final temperature of the mixture is 20.70C and the specific heat of the

container is 0.9kJ/kg×K?

The heat received by the calorimeter

Q = malcalΔT + mwatcwatΔT

=0.4kg(0.9kJ/kg×K)2.7K+1kg(4.18kJ/kg×K)2.7K=(0.972+11.297)kJ=12.269kJ

3

According to the energy conservation law, the lead sample lost the same amount

of heat,

Q = mleadcleadΔT = 1.2kg×80K×clead = 12.269kJ.

Therefore,

clead =12.269kJ/(1.2kg×80K) = 0.128 kJ/kg×K.

Phase Transformations and Latent Heat

Common examples of phase changes are melting (solid to liquid), fusion or

crystallization (liquid to solid), vaporization and boiling (liquid to gas),

condensation (gas or vapor to liquid), etc.

A change in phase at a given pressure occurs at a particular temperature and

continues until the transformation is complete without a change in temperature.

The heat necessary to complete the phase transformation

Q = mL,

where L is called the latent heat.

4

Example 3. How much heat is needed to transform 2kg of ice at 1 atm and

-100C into steam?

First, the ice should be warmed to 00C:

Q1 = mcΔT = 2kg×2.05kJ/K·kg×10K = 41kJ

Second, the ice should be melted (use the latent heat L from the Table):

Q2 = mLf = 2kg×333kJ/kg = 666kJ

The heat is also needed to bring the temperature from 00C to 1000C:

Q3 = mc ΔT = 2kg×4.18kJ/K·kg×100K = 836kJ

Vaporization of all this water requires

Q4 = mLv = 2kg×2.26MJ/kg = 4.52MJ

The overall amount of heat is then

Q = Q1 + Q2 +Q3 + Q4 = 41kJ + 666kJ +836kJ + 4520kJ = 6.063MJ

5

The First Law of Thermodynamics

The temperature of a system can be raised not only by bringing in some heat, but

also by using mechanical energy (the mechanical equivalent of heat), such as,

for example, by rotating a paddle inside water (Joule’s experiment).

The energy conservation law can be written as

ΔEint = Qin + Won

where ΔEint is the change in the internal energy of the system, Qin stands for the

heat brought into the system and Won for the work done on the system.

In the differential form this equation represents the first law of thermodynamics,

dEint = dQin + dWon .

If all internal energy of the gas is just the translational kinetic energy, then

Eint = ��nRT.

The equation becomes more complicated for gases consisting of more complex

molecule with internal vibrational and rotational modes.

6

Work and PV diagram for a gas

� !" ��# = $%�% = &'�% = &�(,� *� ��# = −� !" ��# = −&�(

*� ��# = − , &�(-.

-�

Work can be calculated using the PV diagram for

a gas. For an isobaric compression,

*� ��# = − / &�(-.-�= −& / �(-.-�= −&((1 − (2) = −&Δ(

7

Initial and final point on a PV diagram can be connected by different paths:

Each path corresponds to a different amount of work, − / &�(-.-�

Isometric heating

(at constant volume)

Followed by an isoba-

Ric compression

Isobaric compression

followed by an isomet-

ric heating

Isothermal (T

= const) compres-

sion

In the case of isothermal compression (“slow” compression),

*� ��# = − , &�( = − , �45( �(

-.

-�

-.

-�= −�45 / 6-- = −�45��-.-�

-.-�

= �45�� -�-.

Amount of work on

the gas is different

for each process!

8

Most of engines and refrigerators are based on cyclic processes.

Example 4. Consider a cyclic process A→B→C→D→A below. Let us find a

total work W done on the gas during the cycle and the total amount of heat Q

transferred into it.

In all cyclic processes the total change

in internal energy ΔEint = Q + W = 0.

Segments B→C and D→A are iso-

metric meaning that there is no work,

WBC = WDA = 0.

Process A→B is isobaric expansion and

7 = − , &1�( = −&1((7 − ()-8

-9

Process C→D is isobaric compression and

': = − , &2�( = −&2((: − (')-<

-=

The total work W = ': + : + 7 + 7' = −(&2 − &1)((: − (')

9

and the total heat is Q = ΔEint – W = (&2 − &1)((: − (').

Heat Capacities of Gases

Heat capacities C = Q/ΔT depend on the processes during which they are

measured. CV is the value of heat capacity measured at constant volume V, and

CP – at constant pressure P.

At constant volume W = 0, QV = ΔEint, and

- = �����/�5.

At constant pressure dW = -PdV, dQV = dEint + PdV and

? = 6@ABC6D + &�(/�5.

According to the gas law, V = nRT/P → �(/�5 = nR/P and

? = ( + �4For simple gases, in which the whole internal energy is the kinetic energy of

translational motion, Eint = (3/2)nRT and

CV = (3/2)nR, CP = (5/2)nR.

10

Heat Capacities of Diatomic Gases

A diatomic molecule has two rotation axes,

y and x. Its kinetic energy contains usual

terms describing translational motion and

two rotational terms,

� = E� FGH2 + E

� FGI2 + E� FGJ2 + E

� KHωx2 + E

� KIωy2

Such a molecule has 5 quadratic degrees of freedom and, according to the

equipartition theorem,

Eint =5×E�nRT =

L�nRT

The heat capacity of the gas

- = 6@ABC6D =

L�nR,

? = ( + �4 = (7/2)nR

11

Example 5. 3.00mol of oxygen O2 are heated from 200C to 1400C.

a) What amount of heat is needed if the volume is kept constant?

b) What amount of heat is needed if the pressure is kept constant?

c) How much work does the gas do?

a) Q = CVΔT = (5/2)nRΔT = (5/2)×3mol×8.314J/mol·K×120K

= 7.483kJ

b) Q = CPΔT = (7/2)nRΔT = (7/2)×3mol×8.314J/mol·K×120K

= 10.476kJ

c) In a) the volume does not change and Wa = 0. In b) the volume changes,

and the work, according to the first law, Won = ΔEint – Q

= (5/2)nRΔT - (7/2)nRΔT = - nRΔT = 2.993kJ

In addition to rotational modes, large

molecules also have vibrational modes

which contribute to heat capacity

12

Heat Capacities of Solids

A solid can be modeled as a regular array

of atoms or molecules located near fixed

equilibrium positions and connected to each

other by springs. The energy of a molecule

consists of translational and vibrational

parts:

� = E� FGH2 + E

� FGI2 + E� FGJ2 + E

� �Hx2 + E� �I"2 +

E� �JM2

According to the equipartition theorem, all these six modes contribute equally

to the internal energy and heat capacity (Dulong-Petit law):

Eint = 6 ×E�nRT = 3nRT, C = 3nR, c’ = 3R

The equipartition theorem breaks down

when the quantum effects in rotation

become important.

13

The Quasi-Static Adiabatic Compression of Gases

The adiabatic processes are the processes in which no heat is transferred in or

out of the systems. The system is either well insulated or the process takes place

very quickly. Since Q = 0, the first law becomes

CVdT = 0 – PdV.

Combining this with the gas equation, we get

CVdT = -nRTdV/V or dT/T + (nR/CV)dV/V = 0.

After integration

lnT + (nR/CV)lnV = const or TVnR/CV = const

Since CP – CV = nR, nR/CV = CP/CV – 1 = γ – 1

where

γ = CP /CV

is called the adiabatic constant. Therefore during the adiabatic process,

TVγ-1 = const and PVγ = const.

Since Q = 0, the work done on the gas in an adiabatic compression is

ΔWon = ΔEint = CVΔT ,

Or, using the gas law,

ΔWon = 9-�N(PfVf – PiVi) = (PfVf – PiVi)/(γ-1)

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For monatomic gases γ = CP/CV = 5/3, for diatomic gases γ = 7/5.

Example 6.

a) How much work is needed to pump the air adiabatically into the

1L tire to raise its pressure from 1atm to 583kPa

b) What would be the pressure in the tire after the pump is removed

and the temperature returns to 200C?

a) Since Q = 0, W = CVΔT. The temperature can be found from the alternative

form of the law of adiabatic compression, Tγ/Pγ-1 = const:

W = CVTi[(Pf/Pi)1-1/γ -1]

The final temperature is

Tf = 293K[(583kPa/101.3kPa)1-5/7 – 1] = 483K

and the work

W = CV(483K – 293K) = (5/2)(PfVf/Tf)(483K – 203K) = 634J

b) Since the air in the tire cools at constant volume,

P3 = T3P2/T2 = (293K/483K)2.64atm = 1.6atm

15

Speed of sound waves

The speed of sound wave is usually given by equation

G = :�6��O���P/Q(wait until Chapter 15!) where ρ is the mass density,

Q = �R( = �R

�45&

= R&45 ,

and :��2�!�S2 is the adiabatic bulk modulus

:�6��O���P = −( 6?6- = γP.

In the end, the sound velocity in a gas is

G = T45/R.

16

Review of Chapter 18

Heat – Q - energy transferred due to a temperature difference ΔT

Heat is often measured in calories; 1cal = 4.184J

Heat capacity: C = Q/ ΔT; specific heat: c = C/m; molar specific heat: c’=C/n

Heat capacity at constant volume: CV; at constant pressure: CP = CV + nR > CV

The internal energy of monatomic gas Eint = (3/2)nRT;

diatomic gas Eint = (5/2)nRT

The internal energy determines the heat capacity: CV = dEint/dT

Latent heat is the amount of heat needed for the phase transformation per unit

mass, L = Q/m

The first law of thermodynamics: ΔEint = Qin + Won

17

Isometric process: V = const W = 0

Isobaric process: P = const Won = -PΔV

Isothermal process: T = const Won = nRTln(Vf/Vi)

Adiabatic process: Q = 0 Won = CVΔT

Adiabatic process in ideal gas: Tγ/Pγ-1 = const; TVγ-1 = const; PVγ = const

γ = CP/CV – adiabatic constant

Work Won = -∫PdV

Equipartition theorem: for each quadratic degree of freedom Eint gains E� �45

Dulong-Petit law: the molar specific heat of most solids is c’ =3R

Speed of sound in gases: G = T45/R