Transcript 6. Absorption of Heat

Unit 3
Temperature, Heat, and
the First Law of Thermodynamics
Absorption of Heat
Q  C  T
You heat an object
(add heat to it)
It gets hot
(temperature increases)
Heat Capacity (cal/K, or J/K)
Q  cm  T
Specific Heat (cal/g · K, or J/kg·K)
By definition of heat, the specific heat
of water is
c = 1 cal/g · K
higher than most other substance
Latent Heat
During some phase transitions (i.e. ice - water),
heating does not lead to increase of
temperature until the transition is completed
The thermal energy required for the transition:
Q  mL
Latent Heat (cal/g, or J/kg)
Work 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
Vf
W
 p  dV
Vi
P-V Diagram
Vf
W
 p  dV
Vi
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!!)
Close cycle  W = enclosed area
Same Vi and Vf, different path
different area
different work
The First Law of Thermodynamics
For given initial and final points, Q - W is the
same for all paths.
The First Law of Thermodynamics:
Eint  Q  W
difference of
internal energy
heat added to
the system
work done by
the system
The First Law of Thermodynamics
Eint  Q  W
or
dEint  dQ  dW
or
dEint  dQ  dWon
Infinitesimal
process
work done ON
the system
The change of internal energy is path
independent
The internal energy is a state function
Eint  Q  W


Adiabatic process:
Q=0

Cyclic process:
∆Eint = 0
∆Eint = -W

Q=W
= Area enclosed
by the cycle

Free expansion:
Q = 0, W = 0

Isovolumetric process:
W=0


∆Eint = 0
∆Eint = Q
Heat Transfer Mechanisms

Conduction
– through the materials

Convection
– through the movement of a heated
substance

Radiation
– through emission of electromagnetic field
Conduction
Exchange of kinetic energy
– Between molecules or atoms (insulators)
– By “free electrons” (metals)
Q
dT
H
 kA
t
dx
Rate of heat
transfer
Thermal
conductivity
(W/m K)
Temperature
gradient
Cross-section
Convection
Heat transfer by the movement of a heated
substance.

Natural convection
Result from difference in density

Forced convection
The heated substance is forced to move
Radiation
Radiation:
Pr  AT
Power of
radiation
4
Temperature
Stefan-Boltzmann constant
 5.67051 x 10-8 W/m2K4
Area
Emissivity
Absorption:
Pa 
4
ATenv
Net absorption:
Pn  Pa  Pr 
4
A (Tenv
4
T )
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?
(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
Q  mL
Q  cm  T
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?
(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)
Ti 
Qw  Qb
 T f  873˚C
cc mc
Q  mL
Q  cm  T
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.
Since the process is a complete cycle (beginning and ending in
the same thermodynamic state), ∆Eint = 0 and Q = W,
QAB + QBC + QCA = 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.
Eint  Q  W
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.
(a) The rate at which the heat is conducted along the rod
kA(TH  TC )
H

L
 16 J/s



 401 W/m  K 4.8  10 -4 m 2 100 C
1.2 m
(b) The rate at which the ice melts is
dm H 16 J/s
 
 0.048 g/s
dt
L 333 J/g
Q  mL
H
Q
dT
 kA
t
dx