Heat and the First Law of Thermodynamics

Ch.17 Heat and the First Law of Thermodynamics17.1 Heat as Energy Transfer 17.2 Internal Energy 17.3 Specific Heat 17.4 The First Law of Thermodynamics 17.5 Applying the First Law of Thermodynamics ; Calculating the Work 17.6 Molar Specific Heat for Gases, and the Equipartition of Energy 17.7 Adiabatic Expansion of a Gas The Absorption of Heat by Solids and Liquids 17.8* Heat Transfer; Conduction, Convection, Radiation

17.1 Heat as Energy TransferHeat is the energy that is transferred between one object and another object because of a temperature difference that exists between them.Heat Q , SI: J.

1cal 4.1860J1cal 3.969 10 3 Btu

Environment

“Calorie”

1Cal 1000cal 4186JHeat and work are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of system.

17.2 Internal EnergyInternal Energy U of an Ideal GasThe total energy of a molecule For ideal gages EP 0 The total energy for N molecule

K EPU N NK1 3 2 K mv k T 2 2

For monatomic ideal gagesR k NAM tot N n N0 M mol

3 3 U N kT n R T 2 2

The internal energy U of an ideal gas is a function of the gas temperature only. U U ( T )

17.3 Specific Heat Heat CapacityQ C T C (T f Ti )dQ C dT

SI: J/K.

Specific HeatQ cm T cm(T f Ti )

dQ SI: J/kg.K. c m dTdQ Cm n dT SI: J/mol.K.

Molar Specific HeatQ nC m T nC m (T f Ti )

Molar Mass M = mNA , Avogadro’s Number NA=6.02 1023mol-1 Number of Moles n=Mtot/M

17.4 The First Law of Thermodynamics The quantity (Q – W) depends on the initial and final states (first law of thermodynamics ) U Q W

dU dQ dW (differential form of the first law ) The internal energy U of a system tends to increase if energy is added as heat Q and tends to decrease if energy is lost as work W done by the system.

Q W U U U 2 U1

the work done by a system W>0 the work done on a system W<0

17.5 Applying the First Law of Thermodynamics; Calculating the Work Work in a thermodynamic process with approximately thermal equilibrium(准静态)

dW F ds pAds pdVW dW Vf Vi

A: area of the surface

F

dV

ds F

pdV

p,V dV

p, V

Work on P-V Diagram

W dW The work done by a systemExpansion

Vf

Vi

pdV

The work done on a systemCompression

Vi

Vf

Vi

Vf

Work on P-V Diagram

dW dWigf icf

Wnet pdVL

Vi

Vf

Wighf Wicdf

Cyclical process

Work Done at Constant V and at Constant P The Constant-Volume Process (Isochoric Process)

V constant, WV 0

W

Vf

Vi

pdV

The Constant-Pressure Process (Isobaric Process) Isochoric

P constant, W p P ( V f Vi )

Isobaric

Work Done by an Ideal Gas at Constant TnRT T constant, p VWT Vf Vi

pdV

Vf

Vi

nRT dV V

Isothermpi

WT nRT

ln

Vf Vi

pf

(isothermal process)Pi WT nRT ln PfVi Vf

Example1(H.p.458) One mole of oxygen (ideal gas) expands at a constant temperature T of 310K from Vi = 12L to Vf = 19L. How much work is done by the gas during the expansion? Solution:

WT nRT ln

Vf Vi

19 1 8.31 310 ln 12 1180( J )

Adiabatic ProcessesQ=0

U W

A Graphical Summary of Four Gas Processes

Q1 Q2 Q3 Q4

T 200K ,Isobaric Isothermal

Isochoric

Adiabatic

Some Special Cases of the First Law of Thermodynamics

Q W UCyclical processes

U 0Qnet WnetFree expansions W=Q=0 U 0

Ti T f

Example 2(H.p.440) The figure here shows four paths on a p-V diagram along which a gas can be taken from state i to f. Rank the paths according to (a) the change U, (b) the work W done by the gas, and (c) the energy transferred as heat Q.Solution:

(a) All tie: U= Uf - U i (b) W

Vf

Vi

pdV

W 4> W 3 >W 2> W 1 (c) Q W U Q4> Q 3 >Q 2> Q 1ViVf

Example 3(H.p.442)

For one complete cycle as shown in the p-V diagram here, are (a) U for the gas and (b) the net transferred as heat Q positive, negative, or zero?.

Solution:

(a) Cyclical process : U= 0 (b)

Q W U U= 0

Q W 0

a state A to another state B and back again to A, via state C, as shown by path ABCA in the p-V diagram. (a) Complete the table. (b) Calculate the work done by the system for the complete cycle ABCA. Solution:

Example 4 (p.452-50) A thermodynamic system is taken from

Q W U(a)

U

W AB 0

0

W BC 0

WCA 0 ,

U CA 0