unit 3
DESCRIPTION
Unit 3. Temperature, Heat, and the First Law of Thermodynamics. You heat an object. It gets hot. Heat Capacity (cal/K, or J/K). Specific Heat (cal/g · K, or J/kg·K). Absorption of Heat. (add heat to it). (temperature increases). By definition of heat, the specific heat of water is - PowerPoint PPT PresentationTRANSCRIPT
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