Unit 3Unit 3

Temperature, Heat, and Temperature, Heat, and the First Law of Thermodynamicsthe First Law of Thermodynamics

Absorption of HeatAbsorption of Heat

Q CT

Q cmT

You heat an object It gets hot

Heat Capacity (cal/K, or J/K)

(add heat to it) (temperature increases)

Specific Heat (cal/g · K, or J/kg·K)

By definition of heat, the specific heat of waterwater is

c = 1 cal/g · Kcal/g · K

higher than most other substance

Latent HeatLatent Heat

During some phase transitions (i.e. ice - water), heating does not lead to increase of temperature until the transition is completed

Q m L

Latent Heat (cal/g, or J/kg)

The thermal energy required for the transition:

Work Done During Volume ChangeWork Done During Volume Change

Consider a gas cylinder of piston area A, gas pressure p, and gas volume V

The gas expands, the piston moves by ds, and the volume changes from V to V+dV=V+A·ds

The work done BY the gas: dW=F·ds=A·p·ds=p·dV

The work done BY the gas during the volume change from Vi to Vf

W pdVVi

Vf

P-V DiagramP-V Diagram

W pdVVi

Vf

is the area under the curve in the p-V diagram representing a path from Vi to Vf

(Pay attention to the direction of the path!!)

different area

different work

Same Vi and Vf, different path

Close cycle W = enclosed area

The First Law of ThermodynamicsThe First Law of Thermodynamics

For given initial and final points, Q - W is the same for all paths.

The First Law of Thermodynamics:The First Law of Thermodynamics:

Eint Q W

difference of internal energy

heat added tothe system

work done bythe system

The First Law of ThermodynamicsThe First Law of Thermodynamics

Eint Q W

work done ONthe system

dEint dQ dWonor

dEint dQ dWor

Infinitesimalprocess

The change of internal energyinternal energy is path independent

The internal energyinternal energy is a state function

Adiabatic process:Adiabatic process: Q = 0 ∆Eint = -W

Cyclic process:Cyclic process: ∆Eint = 0 Q = W

= Area enclosedby the cycle

Free expansionFree expansion: : Q = 0, W = 0 ∆Eint = 0

Isovolumetric process:Isovolumetric process:

W = 0 ∆Eint = Q

Eint Q W

Heat Transfer MechanismsHeat Transfer Mechanisms

Conduction– through the materials

Convection– through the movement of a heated

substance

Radiation– through emission of electromagnetic field

ConductionConduction

Exchange of kinetic energy Exchange of kinetic energy – Between molecules or atoms (insulators)Between molecules or atoms (insulators)– By “free electrons” (metals)By “free electrons” (metals)

H Q

t kA

dT

dx

Rate of heattransfer

Temperaturegradient

Cross-sectionThermalconductivity

(W/m K)

ConvectionConvection

Natural convection Natural convection

Result from difference in densityResult from difference in density

Forced convection Forced convection The heated substance is forced to The heated substance is forced to

movemove

Heat transfer by the movement of a heatedHeat transfer by the movement of a heatedsubstance.substance.

RadiationRadiation

Pr AT 4

Power of radiation

Temperature

Stefan-Boltzmann constant5.67051 x 10-8 W/m2K4

Emissivity

Area

Radiation:

Pa ATenv4

Absorption:

Net absorption:

Pn Pa Pr A (Tenv4 T 4 )

(a) The heat transferred to the water of mass ml is:

Qw = cwmw∆T + LVms

= (1 cal/gC˚)(220g)(100˚C-20.0˚C)+(539 cal/g)(5.00 g)= 20.3 kcal

HRW 54E (5th ed.). A 150 g copper bowl contains 220 g of water, both at 20.0˚C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100˚C. (a) How much heat was transferred to the water? (b) How much to the bowl? (c) What was the original temperature of the cylinder?

Q cmT

Q m L

(b) The heat transferred to the bowl is:Qb = cbmb∆T= (0.0923 cal/gC˚)(150g)(100˚C-20.0˚C)= 1.11 kcal

(c) Let it be Ti, then -Qw - Qb = ccmc(Tf-Ti)

CTmc

QQT f

cc

bwi ˚873

HRW 54E (5th ed.). A 150 g copper bowl contains 220 g of water, both at 20.0˚C. A very hot 300 g copper cylinder is dropped into the water, causing the water to boil, with 5.00 g being converted to steam. The final temperature of the system is 100˚C. (a) How much heat was transferred to the water? (b) How much to the bowl? (c) What was the original temperature of the cylinder?

Q cmT

Q m L

Since the process is a complete cycle (beginning and ending inthe same thermodynamic state), ∆Eint = 0 and Q = W,

QCA = W- QAB - QBC = 15.0 J - 20.0 J - 0 = -5.0 J

5.0 J of energy leaves the gas in the form of heat.

QAB + QBC + QCA = W

HRW 75E (5th ed.). Gas within a chamber passes through the cycle shown in Fig. 19-37. Determine the net heat added to the system during process CA if the heat QAB added during process AB is 20.0 J, no heat is transferred during process BC, and the net work dome during the cycle is 15.0 J.

Eint Q W

(b) The rate at which the ice melts is

dm

dt

H

L

16 J/s

333 J/g0.048 g/s

H kA(TH TC )

L

401 W/m K 4.810-4 m2 100 C 1.2 m

16 J/s

(a) The rate at which the heat is conducted along the rod

HRW 84E (5th ed.). A cylindrical copper rod of length 1.2 m and cross-sectional area 4.8 cm2 is insulated to prevent heat loss through its surface. The ends are maintained at a temperature difference of 100˚C by having one end in a water-ice mixture and the other in boiling water and steam. (a) Find the rate at which heat is conducted along the rod. (b) Find the rate at which ice melts at the cold end.

H Q

t kA

dT

dx

Q m L