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Ch 12 Heat
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Updated 5/26/03

PA STANDARDS: a] Analyze energy sources and transfers of heat.
b] Use conservation & entropy to solve energy & heat problems.
c] Apply and analyze energy sources and conversions and their
   relationship to heat and temperature. 

AIR      1.10       ETHANOL     2.45       LITHIUM     3.41       H2SO4      1.54
ALUMINUM 0.903      FREON 11    0.856      MERCURY     0.138      TIN        0.220
AMMONIA  4.60       HELIUM      3.12       METHANOL    2.37       WATER      4.18
BRASS    0.376      GLASS       0.664      NITROGEN    1.039        ICE      2.06
CARBON   0.716      GOLD       0.129       OXYGEN      0.909        STEAM    2.02
CO2      0.775      IRON       0.450       SILVER      0.235      ZINC       0.386
COPPER   0.385      LEAD       0.130       STEEL       0.450

 1    a] Describe and interpret the Food Color Demo.      1
One drop of food color is put into a beaker of cold water and a beaker of hot water.  The drop in cold water falls to the bottom.  The drop in hot water diffuses within 5 min.  The diffusion is caused by the more rapid motion of hot water molecules.  Thus, temp is a measure of how fast the molecules move. 
b] Temperature is the measure of the average internal kinetic energy of the particles (atoms & molecules) in a material.  It does NOT include the external K.E. of the whole object if it is in motion.
c] Heat is the transfer of internal K.E. from one object to another.  This transfer occurs through the collisions of moving surface atoms
(vibrations for solids, translations for fluids). 
Heat is energy in motion.
d] Thermal equilibrium between two objects in contact (A & B) occurs when the transfer from "A" to "B" equals the transfer from "B" to "A".

 2    a] THERMOMETERS reach thermal equilibrium with an object and then measure some temperature dependent property. 
b] LIST 6 EXAMPLES OF HOW THERMOMETERS WORK.
    Thermometer     | Temperature Dependent Property
Liquid in glass     | Volume
Dial (meat)         | Length
Digital             | Electrical Resistance
Thermocouple (oven) | Voltage
Color Strip         | Crystal Angle
Pyrometer (fire)    | Color

Liquid in glass thermometers are limited by the freezing point and boiling point of the liquid.  For example, a mercury thermometer cannot measure temperatures below -39 oC or above 357 oC
c] Calibrate:  to adjust an instrument so that its numbers match those of a standard.
d] To calibrate a thermometer
   1] Put it in ice water and mark 0o on the thermometer
   2] Put it in boiling water and mark 100o on the thermometer
   3] Divide the distance between the marks into 50 equal spaces
      each worth 2o.

 3    a] HEAT GAIN = MASSxSPECIFIC HEAT x /\T          Q      Jules
       Q     =   m x     C        X /\T. b] C = ----- = -----
                                                m /\T    g*oC
  SPECIFIC HEAT CAPACITY [C] is the heat needed to raise 1 g 1 oC
It is a measure of how much heat an object can store.
d] It is a property of matter, and can be used to identify unknowns. 
e] CH2O/CStone = 4.18/0.664 = 6.30 ---> CH2O = 6.30xCStone
f] Implications:  H2O stabalizes atmospheric temperature.  Ex.:  Night vs day temp changes are more extreme in a desert than in an island.


 4    a] 50 g OF SILVER & 50 g OF GOLD ARE HEATED TO 100 oC WHICH METAL WILL CAUSE MORE ICE TO MELT?  EXPLAIN.
Specific Heat is a measure of how much heat an object can store.
The object that stores more heat will melt more ice.
Since silver stores 0.235 J per g (=CAg) while gold stores only
0.129 J per g (= CAu), silver will melt more ice.  CAg/CAu = 0.235/0.129 = 1.82 ==>silver holds 82% more heat & melts 82% more ice
b] WILL AN ICE CUBE LOWER THE TEMP OF A GLASS OF WATER OR A GLASS OF WINE [ETHANOL] FASTER?  EXPLAIN.
A gram of ice can absorb a fixed amount of heat, (80 cal).
For equal masses of melted ice the
heat absorbed by water = heat absorbed by ethanol: 
                Cw/\Tw = Ce/\Te
                 4x10  =  2x20
Thus, the liquid that stores less heat (small C) must have a larger /\T to release the same amount of heat.  Cethanol = 2.45 cools faster.

 5    HOW MUCH HEAT IS ABSORBED BY 250 g OF WATER WHEN IT IS HEATED FROM
10 oC TO 85 oC?  Q = mC/\T--->Q = 250 g*4.18 J/goC*(85-10)oC = 78375 J

 6    A 100 g OF GLASS AT 90oC IS MIXED WITH 100 g OF WATER AT 20 oC
a] DERIVE THE EQUATION FOR THE FINAL TEMPERATURE OF A MIXTURE,
   HEAT LOST BY HOT = HEAT GAINED BY COLD
           Qh       =        Qc
   Mh Ch (Th - Tf) = Mc Cc (Tf - Tc)&nbbsp;  Mh & Mc = the hot & cold mass
                Mh Ch Th + Mc Cc Tc    Ch & Cc = the hot & cold SpHeat
Algebra--->Tf = -------------------    Th & Tc = the hot & cold temp
                   Mh Ch + Mc Cc            Tf = Final Temp
                                                 Weighted Avg
Note:  If both masses are equal:  Mh = Mc
            Ch Th + Cc Tc
       Tf = -------------   a weighted average
               Ch + Cc  
Note:  If both materials are the same:  Ch = Cc
            Mh Th + Mc Tc
       Tf = -------------   a weighted average
               Mh + Mc  
Note:  If both masses are equal & both materials are the same.
            Th + Tc
       Tf = -------&   an average
               2
b] RANGE OF POSSIBLE ANSWERS:  Th > Tf > Tc
c] FINAL TEMPERATURE OF THE MIXTURE.  To avoid mistakes.
   1] Identify the given   [2] Use the most general equation
Ex:  PRACT 1:  600 g CARBON AT 70oC IS PLACED IN 500 g OF H2O AT 30oC
Hot:  Mh = 600 g, Th = 70oC, Ch = 0.716 (Carbon)
Cold: Mc = 500 g, Tc = 30oC, Cc = 4.18 (water)     

     Mh Ch Th + Mc Cc Tc   600*0.716*70 + 500*4.18*20   71,872
Tf = ------------------- = -------------------------- = ------ = 28.53
        Mh Ch + Mc Cc         600*0.716 + 500*4.18       2,519
Note:  to avoid errors divide the calculation into three steps
       (top, bottom, divide).
d] ASSUMPTION:  No heat lost to or gained from the surroundings.

 7    a] DERIVE THE EQUATION FOR THE SPECIFIC HEAT OF A MATERIAL.
Heat Lost by Hot = Heat Gained by Cold
         Qh          =      Qc
   Mh Ch (Th - Tf)  = Mc Cc (Tf - Tc)   [Divide by Mh (Th - Tf)
        Mc Cc (Tf - Tc)  
   Ch = ---------------
         Mh (Th - Tf) 
b]  A 50 g MASS AT 100oC IS PLACED IN 100 g OF WATER AT 10oC, Tf = 13.87oC.  FIND THE SPECIFIC HEAT & THE MATERIAL. 
     Mc Cc (Tf - Tc)   100*4.18*(13.87 - 10)   1617.66
Ch = --------------- = --------------------- = ------- = 0.376
      Mh (Th - Tf)      50*(100 - 13.87)>       4306.50
Note:  to avoid errors divide the calculation into three steps
       (top, bottom, divide).
ANS:  Ch = 0.376-->brass (see table)

8     a] THE 5 TYPES OF HEAT TRANSFER.
RADIATION--Heat transfer by infrared light.
Ex:  Sun warming the earth, toaster.
Note:  Microwave Ovens transfer energy not heat, they warm by vibrating water and don't heat things without water, i.e. styrofoam.
NATURAL CONVECTION--Heat transfer by the flow of a fluid caused by a hot or cold spot.  Ex:  Heat flow around a flame or radiator.
FORCED CONVECTION--Heat transfer by the flow of a fluid caused by a pump or a fan.  Ex:  hot water flowing through a radiator.
CONDUCTION--heat transfer by direct contact.
Ex:  Heat flow through a metal spoon left in a cup of hot coffee.
EVAPORATION:  heat transfer by a liquid to gas phase change.
Liquids absorb heat to evaporate:  Ex:  sweating, each gram of sweat absorbs 2,260 J from your body to evaporate.
b] See board.


Heat Theory

Updated 5/26/03
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CLASSIFY 8 TYPES OF HEAT ENGINES.
                         RECIPROCATING      ROTARY
EXTERNAL COMBUSTION      STEAM     [1]   STEAM TURBANS [5]
INTERNAL COMBUSTION      GASOLINE        ROTARY        [6]
                          4 CYCLE   [2]   GAS TURBAN    [7]
                          2 CYCLE   [3]
                          DIESEL    [4]   ROCKET        [8]

KINETIC THEORY:  A] ALL MATTER CONSISTS OF MINUTE PARTICLES WHICH ARE IN CONSTANT MOTION.  DUE TO THEIR MOTION THEY HAVE KINETIC ENERGY.  B] THESE PARTICLES MUST ATTRACT ONE ANOTHER.  OTHERWISE, NO SOLID OR LIQUID EXIST

IT IS IMPOSSIBLE AND IMPRACTICAL TO TRY TO FIND THE TOTAL INTERNAL ENERGY OF ANY GIVEN OBJECT.  HOWEVER, STUDYING THE CHANGE IN INTERNAL ENERGY WHEN ENERGY IS ADDED OR REMOVED FROM AN OBJECT WILL YIELD GREAT SUCCESS IN SOLVING PROBLEMS.  THIS TRANSFER OF ENERGY IS CALLED HEAT ENERGY.

SYSTEM--ANY WELL-DEFINED COLLECTION OF OBJECTS:  A SINGLE SPRING, OR A MAGNET, OR AN IDEAL GAS IN A CONTAINER, OR SOME COMPLICATED GADGET WITH MANY INTERNAL PARTS.

COLD--A TERM THAT DESCRIBES THE HEAT ENERGY OF ONE OBJECT RELATIVE TO THE HEAT ENERGY OF ANOTHER.  COLD DOES NOT FLOW, HEAT FLOWS.

 1    DEFINE THERMAL ENERGY, HEAT, TEMPERATURE, AND ABSOLUTE ZERO
THERMAL ENERGY--THE TOTAL INTERNAL K.E. & P.E. OF A SYSTEM
HEAT--A MEASURE OF THE INTERNAL K.E. TRANSFERRED FROM ONE OBJECT TO ANOTHER
TEMPERATURE--THE MEASURE OF AN OBJECT'S AVG INTERNAL KINETIC ENERGY.
ABSOLUTE ZERO [-273 oC, -460 oF]--ALL TRANSLATIONAL ENERGY DISAPPEARS BUT A GREAT RESERVOIR OF P.E. AND A SMALL AMOUNT OF VIBRATIONAL ENERGY REMAIN.  [AVERAGE K.E. = 0, GAS VOLUME = 0]. 

 2    NAME THE 4 TEMPERATURE SCALES AND THEIR STANDARDS.
WATER.  C = 5/9[F - 32], F = 9/5 C + 32; AN IDEAL GAS. 
K = oC + 273, OR oC = K - 273           R = oF + 460, OR oF = R - 460

 4a   HOW DO THERMOMETERS WORK?  THEY REACH EQUILIBRIUM WITH THE OBJECT THEY CONTACT, AND THEN MEASURE SOME TEMPERATURE DEPENDENT PROPERTY. 
b     LIST 5 TYPES OF THERMOMETERS & THE PROPERTY USED TO MEASURE TEMP.
LIQUID IN GLASS--MERCURY OR ALCOHOL -- VOLUME. 
THERMOCOUPLES [OVEN THERMOSTATS] -- VOLTAGE;
RESISTANCE [DIGITAL] -- ELECTRICAL RESISTANCE [-259 TO 962 oC]
BIMETALIC [HOUSE THERMOSTATES] -- DISTANCE;
COLOR STRIP [FISH TANK] -- CRYSTAL ORIENTATION
HELIUM -- VAPOR PRESSURE [VERY LOW TEMPERATURES]
OPTICAL PYROMTER-- COLOR OF GLOWING SUBSTANCES [VERY HIGH TEMPS]
c     WHAT LIMITS THE USE OF Hg THERMOMETER? BP & FP OF Hg, -39oC TO 357 oC

11    DEFINE AND GIVE ONE EXAMPLE OF THE 4 TYPES OF HEAT TRANSFER.
RADIATION--HEAT TRANSFER BY INFRARED LIGHT.
NATURAL CONVECTION--HEAT TRANSFER BY THE FLOW OF A FLUID CAUSED BY A HOT OR COLD SPOT.
FORCED CONVECTION--HEAT TRANSFER BY THE FLOW OF A FLUID CAUSED BY A PUMP OR A FAN.
CONDUCTION--HEAT TRANSFER BY DIRECT CONTACT.

12    WRITE THE EQUATION FOR THE RATE OF HEAT TRANSFER & DEFINE THE VARIABLES.

2     USE THE EQUATION FOR HEAT GAIN [Q = mC/\T] TO EXPLAIN YOUR ANSWERS TO:
a] FIFTY GRAMS OF METAL X AND FIFTY GRAMS OF METAL Y ARE HEATED TO THE
   SAME TEMPERATURE.  THE PIECES OF METAL ARE PLACED ON A BLOCK OF ICE
   WHICH METAL WILL CAUSE MORE ICE TO MELT?  SINCE m & /\T ARE EQUAL THE METAL THAT HOLDS MORE HEAT WILL BE THE ONE WITH THE GREATER SPECIFIC HEAT CONTENT [C].
   COMPARE SILVER & GOLD.  Csilver/Cgold = 0.235/0.129 = 1.822; thus silver holds 82.2% more heat.
b] WILL AN ICE CUBE LOWER THE TEMPERATURE OF A GLASS OF WATER OR A
   GLASS CONTAINING AN ALCOHOLIC BEVERAGE FASTER?  WHY?
It takes 4.18 J to lower a gram of water, and 2.45 J.
A gram of ice can absorb a fixed amount of heat, Ex 100 J. 
Thus it will lower a g of H2O /\T = Q/mC = 100 J/(1g 4.18) = 19.31oC
& a g of ETHANOL /\T = Q/mC = 100 J/(1g 2.45) = 28.99 oC


Heat Engines & the 2nd Law

Updated 5/26/03
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ADDITIONAL QUESTIONS

      STATE THE FIRST LAW OF THERMODYNAMICS.  Q = /\U + W:  Q = HEAT SUPPLIED TO A SYSTEM, /\U = CHANGE IN INTERNAL E, W = WORK DONE BY SYS

      STATE & GIVE AN EXAMPLE OF THE SECOND LAW OF THERMODYNAMICS.

      HEAT ENERGY IS ___ WHEN CHANGED TO ANOTHER FORM.

      DOES TEMPERATURE FLOW FROM A HOT OBJECT TO A COLD ONE?  EXPLAIN.

      SKETCH & LABEL THE PARTICLE FRACTION vs K.E. FOR 2 TEMPS.  [CHEM 324]

      ___ NEVER DECREASES IN NATURAL PROCESSES.

      IS ENERGY LOST WHEN A FUEL IS BURNED?  EXPLAIN

      ___ NEVER DECREASES IN NATURAL PROCESSES.

      A ___ ENGINE CONTINUOUSLY CONVERTS THERMAL ENERGY TO MECHANICAL ENERGY

      A ___ USES MECHANICAL ENERGY TO TRANSFER HEAT FROM AN AREA OF LOWER TEMP TO AN AREA OF HIGHER TEMP.

      IF YOU PLACE IDENTICAL UNCOVERED PAILS OF HOT AND COLD WATER OUTDOORS ON A DRY SIDEWALK ON A VERY COLD DAY, THE HOT WATER USUALLY STARTS TO FREEZE FIRST.  EXPLAIN.  WHAT WOULD HAPPEN IF THE PAILS WERE COVERED?

      EXPLAIN WHY FRUIT GROWERS SPRAY THEIR TREES WITH WATER, WHEN FROST IS EXPECTED, TO PROTECT THE FRUIT FROM FREEZING.

      WHY DOES WATER IN A CANTEEN STAY COOLER IF IT HAS A CANVAS COVER THAT IS KEPT WET?

14    a      DEFINE HEAT ENGINE & HEAT PUMP.
HEAT ENGINES:  ANY DEVICE OPERATING THROUGH A CYCLE, CONVERTS [SOME] HEAT TO WORK & DISCRADS THE REMAINDER INTO A COLD RESERVOIR.
HEAT PUMP:  A HEAT ENGINE RUN BACKWARDS--CONVERTS WORK INTO HEAT

THE 2nd LAW OF THERMODYNAMICS--NATURAL PROCESSES GO IN THE DIRECTION THAT INCREASES RANDOMNESS--ENTROPY.  EX:  HEAT FLOWS SPONTANEOUSLY FROM A HOTTER TO A COLDER OBJECT BUT NOT THE REVERSE.  EVEN THE MOST IDEAL ENGINE WILL GENERATE SOME WASTE HEAT.

TO USE THE ENERGY STORED IN CHEMICAL OR NUCLEAR FUELS, SUCH AS COAL, OIL, NATURAL GAS, OR URANIUM, ONE TYPICALLY FIRST CONVERTS THE FUEL'S POTENTIAL ENERGY INTO THERMAL ENERGY AND THEN CONVERTS SOME OF THE THERMAL ENERGY TO WORK.  CHEMICAL OR NUCLEAR POTENTIAL ENERGY CAN BE CONVERTED TO THERMAL ENERGY WITH 100% EFFICIENCY.  HOWEVER, A PROCESS IN WHICH HEAT IS CONVERTED TO WORK THROUGH THE USE OF A HEAT ENGINE OPERATING OVER A CYCLE CAN NEVER HAVE 100% EFFICIENCY.

A HEAT ENGINE IS DEFINED AS ANY DEVICE THAT IN OPERATING THROUGH A CYCLE, CONVERTS SOME HEAT TO WORK AND DISCARDS THE REMAINDER INTO A COLD RESERVOIR.  A HEAT ENGINE IS ALWAYS RETURNED TO ITS INITIAL STATE IN A CYCLE.  THE HEAT ENGINES WE CONSIDER ARE IDEALIZED:  THEY HAVE NO FRICTION.  BUT EVEN IF NO ENERGY IS LOST TO FRICTION, AN ENGINE CAN NEVER BE PERFECTLY EFFICIENT.  THE REASONS ARE FAR MORE FUNDAMENTAL.

EFFICIENCY = WORK OUT/HEAT IN  =  W/Q
W = WORK OUT PER CYCLE AND Q = THE HEAT IN PER CYCLE.  WHEN WE APPLY THE FIRST LAW OF THERMODYNAMICS TO ONE CYCLE, WE GET
Q = /\U + W
Qh - Qc = 0 + W; Qh = HEAT IN AT THE HIGH TEMP, Qc = HEAT OUT AT COLD TEMP.  THIS SAYS THAT THE WORK OUT EQUALS THE NET HEAT IN.  IN COMPLETING ONE CYCLE, THE SYSTEM RETURNS TO ITS INITIAL STATE.  THEREFORE, /\U = 0.  USING THIS RESULT WE HAVE:  EFFICIENCY = (Qh - Qc)/Qh = 1 - Qc/Qh.

THIS SHOWS THAT AN ENGINE CAN HAVE 100% EFFICIENCY ONLY IF Qc =0.  AN ENGINE CAN BE PERFECTLY EFFICIENT ONLY IF NO THERMAL ENERGY IS EXHAUSTED TO THE COLD RESERVOIR.  THIS IS IMPOSSIBLE.

THE REASON?  LIKE ANY OTHER FUNDAMENTAL LAW IN PHYSICS, IT IS CONFIRMED BY THE CIRCUMSTANCES THAT NO EXCEPTION TO IT HAVE EVER BEEN FOUND.

THE SECOND LAW OF THERMODYNAMICS CAN BE STATED AS:
NO HEAT ENGINE, REVERSIBLE OR IRREVERSIBLE, OPERATING IN A CYCLE, CAN TAKE IN THERMAL ENERGY FROM ITS SURROUNDINGS AND CONVERT IT ALL TO WORK.
LORD KELVIN 1851.

A HEAT ENGINE RUN IN REVERSE IS A HEAT PUMP.  WORK W IS DONE ON A SYSTEM, HEAT Qc IS EXTRACTED FROM THE LOW-TEMP RESERVOIR, AND HEAT Qh IS EXHAUSTED TO THE HIGH-TEMP RESERVOIR.  THE NET EFFECT IS THAT HEAT IS PUMPED FROM THE LOW TO THE HIGH RESERVOIR.  A HEAT PUMP IS EFFECTIVELY A REFRIGERATOR.

.C.*PHASE CHANGE

1     CAN YOU ADD THERMAL ENERGY TO AN OBJECT WITHOUT INCREASING ITS TEMPERATURE?  EXPLAIN

2     IF YOU PLACE IDENTICAL UNCOVERED PAILS OF HOT AND COLD WATER OUTDOORS ON A DRY SIDEWALK ON A VERY COLD DAY, THE HOT WATER USUALLY STARTS TO FREEZE FIRST. 
a] EXPLAIN; b] WHAT WOULD HAVE HAPPENED HAD THE PAILS BEEN COVERED?

3     ARE THE COILS OF AN AIR CONDITIONER THAT ARE INSIDE THE HOUSE THE LOCATION OF VAPORIZATION OR CONDENSATION OF THE FREON?  EXPLAIN.

4      EXPLAIN WHY FRUIT GROWERS SPRAY THEIR TREES WITH WATER, WHEN FROST IS EXPECTED, TO PROTECT THE FRUIT FROM FREEZING.

5     WHY DOES WATER IN A CANTEEN STAY COOLER IF IT HAS A CANVAS COVER THAT IS KEPT WET?

6     WHY DO SOME FUSION REACTORS LIQUID LITHIUM AS A COOLANT

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