CHAPTER 3: TEMPERATURE AND HEAT by hR2mHk

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									ESM 1224, SEMESTER 3, 2006/2007                                           CHAPTER 3



               CHAPTER 3: TEMPERATURE AND HEAT

 3.1 Temperature and Thermometers
 Temperature is a physical property of all bodies and is a relative measure or
    indication of hotness and coldness of objects.
   Thermometers are devices used to measure the temperature of a system.
   Three common scales of a temperature measurement:
                1. Celsius
                2. Fahrenheit
                3. Kelvin

   Table shows a comparison of the Celsius, Fahrenheit and Kelvin temperature
    scale
                           Celsius        Fahrenheit       Kelvin
                            scale            scale          scale
           Steam point      100 C           212 F       373.15 K
             Ice point       0 C             32 F       273.15 K

   On the Celsius temperature scale, there are 100 equal divisions between the
    ice point (0C) and the steam point (100C).
   On the Fahrenheit temperature scale, there are 180 equal divisions between
    the ice point (32F) and the steam point (212F).
   The relationship between Celsius, TC and Fahrenheit, TF temperature scale is
                                      9
                                  TF  TC  32
                                      5
3.2    The Kelvin Temperature Scale
   For scientific work, the Kelvin temperature scale is the scale of choice.
   As shown in the table above, the ice point (0C) occurs at 273.15K on the
    Kelvin scale.
   Thus, the Kelvin temperature, T and the Celsius temperature are related by,
                            T  TC  273 .15
   Thus, the Kelvin temperature, T and the Fahrenheit temperature are related
    by,
                            TF  9 T  273.15  32.0
                                 5

   The number 273.15 in the equation is an experimental result, obtained in
    studies that utilize a gas-based thermometer.
   Gas thermometer is a standard instrument because its measurement is
    independent from any type of gas.
   One type of gas thermometer is the constant-volume gas thermometer.
   The physical property used in a constant-volume gas thermometer is the
    pressure variation with temperature of a fixed volume of gas.
   A constant-volume gas thermometer consists of a gas-filled bulb to which a
    pressure gauge is attached, as in figure below.




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     Figure shows a constant-volume gas thermometer




   The basic idea of this thermometer is that the volume of gas is held constant
    by raising or lowering the right column of the U tube manometer in order to
    keep the mercury level in the left column at the same reference level.
   With the mercury level that can be adjusted, the gas occupies a constant
    volume and its pressure is related to

                                  P2= P1 + gh
       where  is the density of mercury
             P1 = 0 Pa (evacuated)
             P2 = Pgas

   As the temperature changes, the pressure changes. The temperature
    increases when a gas confined to a fixed volume is heated and decreases
    when the gas is cooled.
   If the results are plotted on a pressure versus temperature graph, a straight
    line is obtained as in figure below.




        Figure shows a plot of absolute pressure versus temperature for a low
         density gas at constant volume

   If the straight line is extended to lower temperatures where the pressure is
    zero, the line crosses the temperature axis at -273.15C. This is the location
    of absolute zero.




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EXERCISE 1
(a) “Room temperature” is often taken to be 68°F. What is this on the Celsius
     scale?
(b) The temperature of the filament in a lightbulb is about 1800°C. What is this on
     the Fahrenheit scale?

SOLUTION
(a)      T  C  T  F  32  68  32  20 C
             o     5
                   9
                           o
                                 
                                     5
                                     9
                                                   o



(b)      T  F   T  C   32  1800   32  3300 F
             o    9
                  5
                       o         9
                                 5
                                                       o




EXERCISE 2
What’s your normal body temperature? It may not be 98.6F, the oft-quoted
average that was determined in the nineteenth century. What is the difference
between these averages expressed in Celsius degrees?

SOLUTION
The difference between these two averages, expressed in Fahrenheit degrees, is
                                 98.6 F  98.2 F  0.4 F
                       9
Since 1 C° is equal to   F°, we can make the following conversion
                       5
                                         1 C 
                               (0.4 F)             0.2 C
                                        (9/5) F

3.2      Thermal Expansion
     When the temperature of a body increases, the average distance between the
      atoms increases and the volume of the body also increases.
     The phenomenon is known as thermal expansion.
     Thermal expansion is a physical phenomenon in which the increment in
      temperature can cause substances to expand.

 Linear Expansion
     Figure below illustrates the linear expansion of a rod whose length is Lo when
      the temperature is To.




          Figure shows when the temperature of a rod is raised by T, the length
           of the rod increases by L

     For linear expansion, an object of length Lo experiences a change L in length
      when the temperature changes by T

                  L  L0 T where  is the coefficient of linear expansion
             OR   L – Lo =  Lo T
             OR   L  L0 (1  T ) (in terms of final length)


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 Thermal stress
   For an object held rigidly in place, a thermal stress can occur when the object
    attempt to expand or contract.
   Thus, the force / stress needed to prevent a solid object from expanding must
    be strong enough to counteract any change in length due to a change in
    temperature.

                                      F    L   L T
                           Stress      Y    Y 0     YT
                                      A    L0     L0

EXERCISE 3
An aluminum bar has the desired length when at 15°C. How much stress is
required to keep it at this length if the temperature increases to 35°C?

SOLUTION
The thermal stress must compensate for the thermal expansion.                 Y is Young’s
modulus for the aluminum.
                       
Stress  F A  Y T  25  106 Co     70 10
                                               9
                                                   N m2   35 C  15 C  3.5 10
                                                              o     o                7
                                                                                         N m2

EXERCISE 4
A steel beam is used in the roadbed of a bridge. The beam is mounted between
two concrete supports when the temperature is 23C, with no room provided for
thermal expansion. What compressional stress must the concrete supports apply
to each end of the beam, if they are to keep the beam from expanding when the
temperature rises to 42C?

SOLUTION
         F    L     Lo T
           Y    Y          Y T
Stress = A    Lo       Lo
         YT  (2 x1011 )(12 x106 )(19)  4.6 x107 N / m2

 The Bimetallic Strip
   Each substance has its own characteristic coefficient of expansion.
   For example, brass rod expands more than the steel rod because brass has a
    larger coefficient of expansion than steel.
   A simple device that uses this principle is a bimetallic strip.
   A bimetallic strip is made from two thin strips of metal that have different
    coefficients of linear expansion as shown in figure below




        Figure shows: (a) A bimetallic strip and how it behaves when (b)
         heated and (c) cooled


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 APPLICATION
   Bimetallic strips are frequently used as adjustable automatic switches in
    electrical appliances.
   Refer figure below.




 In part (a), the electricity passes through the heating coil that heat the water.
  The electricity can flow because bimetallic strip touches the contact on the
  strength adjustment knob, thus providing a continuous path of the electricity.

 In part (b), when the bimetallic strip gets hot enough to bend away, the
  contacts separate and the electricity stops because it is no longer has a
  continuous path along which to flow and the brewing cycle is shut off.

 NOTE:
    Expansion of holes-Let’s say we have a hole in piece of solid material. When
    heated, the material will expand, but what about the hole???  The hole also
    expands at the same rate and at the same direction (outwards)-refer to the
    diagram.




                  (a) Unheated
                                                (b) Heated –
                                                 dotted line
                                              represents after
                                                   heated
 Volume Thermal Expansion
   For volume expansion, the change V in the volume of an object of volume Vo
    is given by V  3V0 T or   V  V0 T
   where  is the coefficient of volume expansion and is equal to 3

                           = 3         (liquids expands more than solids)




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EXERCISE 5
A rod made from a particular alloy is heated from 25 0C to the boiling point of
water. Its length increases by 8.47  10-4 m. The rod is then cooled from 250C to
the freezing point of water. By how much does the rod shrink?

SOLUTION
Assuming that the rod expands linearly with heat, we first calculate the quantity
L/ T using the data given in the problem.
L      8.47  10–4 m               –5
                               1.13  10   m/C
T        100.0 C – 25.0 C

Therefore, when the rod is cooled from 25.0 °C, it will shrink by
L  (1.13  10 –5 m/C) T
      (1.13  10 –5 m/C) (0.00 C – 25.0  C)  –2.82  10 –4 m

EXERCISE 6
A cylinder of diameter 1.00000cm at 30C is to be slid into a hole in a steel plate.
The hole has a diameter of 0.99970cm at 30C. To what temperature must the
plate be heated? For steel,  = 1.2 x 10-5C-1

SOLUTION
L= LoT
      L
T=
      Lo
L= 1  10-2 – 9.997  10-3 = 3  10-6m
= 1.2  10-5C-1
Lo= 9.997  10-3m
             3  106
T=                             = 25C
     1.2  105 (9.997  103 )
The temperature of the plate must be = 30 + 25 = 55C

EXERCISE 7
The brass bar and the aluminum bar in the drawing are attached to an immovable
wall. At 28C the air gap between the rods is 1.3 x 10 -3m. At what temperature
will the gap be closed?




SOLUTION
L = L0T gives for the expansion of the aluminum
LA = ALAT

and for the expansion of the brass
LB = BLBT

Taking the coefficients of thermal expansion for aluminum and brass from table
adding Equations (1) and (2), and solving for T give




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         L A  L B                              1.3 10 3 m
T                                                                                     21C
        A L A   BL B       23  10 6 C 1 1.0 m  19  10 6 C 1 2.0 m
The desired temperature is then
T = 28 °C + 21 C° =        49 C

EXERCISE 8
A lead object and a quartz object each have the same initial volume. The volume
of each increases by the same amount because the temperature increases. If the
temperature of the lead object increases by 4.0C., by how much does the
temperature of the quartz increase?

SOLUTION
Recognizing that the lead and quartz objects experience the same change in
volume and expressing that change with Equation 12.3, we have

                               LeadV0TLead  QuartzV0TQuartz
                                  VLead           VQuartz
In this result V0 is the initial volume of each object. Solving for TQuartz and
taking values for the coefficients of volume expansion for lead and quartz from
Table gives
                                                       1
                                                         
                          Lead TLead 87 10  C    4.0 C 
                                               6
            TQuartz                                       1
                                                                    230 C
                             Quartz        1.5 106  C 

EXERCISE 9
A glass is filled “to the mark” with 50.00cm3 of mercury at 18C. If the flask and
its contents are heated to 38C, how much mercury will be above the mark? glass
= 8.5 x 10-6C-1 and βmercury = 182 x 10-6C-1

SOLUTION
Both of glass and mercury will expand
Vmercury= βmercuryVoT
= 182  10-6 (5  10-5) (38 – 18)
= 1.82  10-7m3

Vglass= βglassVoT
= 3glassVoT
= (3  8.5  10-6) (5  10—5) (38 -18) = 2.55  10-8m3

Volume of mercury above mark = Vmercury - Vglass
= 1.82  10-7 – 2.55  10-8 = 1.565  10-7m3

EXERCISE 10
Suppose that the steel gas tank in your car is completely filled when the
temperature is 170C. How many gallons will spill out of the twenty gallon tank
when the temperature rises to 35 0C?




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SOLUTION
Both the gasoline and the tank expand as the temperature increases. The
coefficients of volumetric expansion  g and s for gasoline and steel are available
in Table.
The volume expansion of the gasoline is


                                           
         Vg   gV0 T  950 10 –6 C 1 (20.0 gal)(18 C)  0.34 gal
While the volume of the steel tank expands by an amount
          V   V T   10 C 
            s    s 0       36     –6      1
                                           (20.0 gal)(18 C)  0.013 gal
                                               Vg  Vs  0.33 gal
The amount of gasoline which spills out is 

3.5    Heat and Internal Energy

   Heat is defined as the energy that flows from a higher temperature
    object to a lower temperature object because of the difference in
    temperatures. (Refer figure)
   The SI unit is Joule , J
   Heat is sometimes measured with a unit called the kilocalorie (kcal)

   The conversion factor between kilocalories and joule is known as the
    mechanical equivalent of heat:
                1 kcal = 4186 joule  or      1 cal = 4.186 joule

   Internal energy of a substance is the sum of the kinetic energy (due to the
    random motion of the molecules), the potential energy (due to forces that act
    between the atoms of the molecule and between molecules) and other kinds
    of energy that the molecules of the substance have.




 Figure shows how heat as energy is transit from hot to cold.
      (a) Heat flows from the hotter coffee to the colder hand.
      (b) Heat flows from the warmer hand to the colder glass of ice water.




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3.6    Heat and Temperature Change: Specific Heat Capacity

 Solids and Liquids
   Greater amounts of heat are needed to raise the temperature of solids or
    liquids to higher values
   Similar concepts apply when the temperature is lowered, except that heat
    must be removed.
   The heat, Q that must be applied or removed to change the temperature of a
    substance of mass, m by an amount T is

                                     Q  mcT
               where c is the specific heat capacity of the substance.

   Unit for specific heat capacity: J / (kg C)

 Calorimetry
 To determine the specific heat of a solid or liquid, we conduct the following
  steps:
     i. raise the temperature of the substance to some value
     ii. place it in a vessel containing cold water of known mass and
          temperature
     iii. measure the temperature of the combination after the equilibrium is
          reached.

   Vessels having this property are called calorimeters and analysis performed
    using such vessels is called calorimetry.

 The principle of conservation of energy for perfectly insulated container
      requires that heat lost by warmer materials equals heat gained by
      cooler materials. (That is if there is no heat loss to the external
      surroundings)

      From the principle of conservation of energy for the isolated system,
        heat loss by the hotter object = heat gained by the cooler object.

                                         Q  0
                                      Q gained  Qlost
                             [mc (T )] gained  [mc (T )] lost

       **When calculating heat contributions, always write any temperature
       changes as the higher minus the lower temperature.

EXERCISE 11
A 0.4kg iron horseshoe that is initially at 500C is dropped into a bucket
containing 20kg of water at 22C. What is the final equilibrium temperature?
Neglect any energy transfer to or from the surroundings

SOLUTION
- Qhot = Qcold
(mCT)iron = (mCT)water
(0.4) (452) (500 – Tf) = (20) (4186) (Tf – 22)
(90400 – 180.8Tf) = (837201Tf – 1341840)
Tf= 39.3C


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EXERCISE 12
An aluminum cup contains 225g of water and a 40g copper stirrer, all at 27C. A
400g sample of silver at an initial temperature of 87C is placed in the water. The
stirrer is used to stir the mixture gently until it reaches its final equilibrium
temperature of 32. Calculate the mass of the aluminum cup.

SOLUTION
The total energy absorbed by the cup, stirrer and water equals the energy given
up by the silver sample. Thus,
(mcup cal + mstirrer ccopper + mwater cwater) (Twater) = (mcT)silver
                             Tsilver
mcup = [ ( msilver csilver            - mstirrerccopper - mwater ) ]
                             Twater
            1                    87  32 
mcup =            [ (0.4) (235)           - 0.04 (387) – 0.225
         9  10                  32  27 
                2

mcup = 0.085kg


EXERCISE 13
An engineer wishes to determine the specific heat of a new alloy. A 0.15kg
sample of the alloy is heated to 540C. It is then quickly placed in 400g of water
at 10.0C, which is contained in a 200g aluminum calorimeter cup. The final
temperature of the mixture is 30.5C. Calculate the specific heat capacity of the
alloy. (The specific heats of water and aluminum are 4186J/kg C and 900J/kgC)

SOLUTION
(heat lost by sample) = (heat gained by water) + (heat gained by calorimeter
cup)
ms cs Ts  mwcw Tw  mcal ccal Tcal
(0.15)cs (540  30.5)  (0.4)(4186)(30.5 10)  (0.2)(900)(30.5  10)
76.4cs  34300  3700
cs  500 J / kg C

EXERCISE 14
A thermometer has a mass of 31.0 g, a specific heat capacity, c = 815 J/kgC and
a temperature of 12.0C. It is immersed in 119 g of water and the final
temperature of the water and thermometer is 41.5C. What was the temperature
of the water before the insertion of the thermometer?

SOLUTION
Since there is no heat lost or gained by the system, the heat lost by the water in
cooling down must be equal to the heat gained by the thermometer in warming
up. The heat Q lost or gained by a substance is given by Equation as Q = cmT,
where c is the specific heat capacity, m is the mass, and T is the change in
temperature. Thus, we have that
                               cH O mH O TH O  ctherm mtherm Ttherm
                                 2       2        2

                                 Heat lost by water   Heat gained by thermometer
We can use this equation to find the temperature of the water before the
insertion of the thermometer.
Solving the equation above for TH 2 O and using the value of cH O from Table we
                                                                                   2

have



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                     ctherm mtherm Ttherm
           TH O 
              2
                          cH O mH O
                            2      2



                   815 J/  kg  C    31.0 g  41.5 C  12.0 C 
                                                                       1.50 C
                               4186 J/  kg  C   119 g 
                                                    
        The temperature of the water before the insertion of the thermometer was
        T  41.5 C  1.50 C = 43.0 C

3.7     Heat and Phase Change: Latent Heat

   In some situations, the transfer of energy does not result in a change in
    temperature.
   This occurred when the substance change from one form to another, which
    referred to as a phase change.
   Figure below summarizes the three familiar phases of matter (solid, liquid and
    gas) and the phase changes that can occur between any two of them

                                 melting                 evaporating

                  solid                         liquid                       gas

                                  freezing                condensing


                                             subliming

                                solid                       gas

                                             condensing


       The heat, Q that must be supplied or removed to change the phase of a
        mass, m of a substance is
                                             Q  mL
        where L is the latent heat of the substance. (Unit: J/kg)SI unit of L=J / kg

       Heat of fusion: the heat required to change 1.0 kg of substance from the
        solid to the liquid state.

       Heat of vaporization: the heat required to change a substance from the
        liquid to the vapor phase.

     Figure below displays a graph that indicates what typically happens when
      heat is added to a material that changes phase.




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    The graph shows the way temperature of water changes as heat is added,
        starting with ice at -40C. The pressure is atmospheric pressure.




      a) From -40C to 0C: no phase change occurs. The temperature of
         the ice changes as heat is added. Amount of energy added is Q =
         mciceT.

      b) At 0C: no temperature change occurs. the ice-water mixture
         remains at this temperature even though energy is being added until
         all the ice melts (phase change occurs) to become water at 0C. The
         energy required to melt is Q = mLf

      c) From 0C to 100C: no phase change occurs. The energy added to
         the water is used to increase its temperature. The amount of energy
         necessary to increase the temperature from 0C to 100C is Q =
         mcwaterT.

      d) At 100C: another phase change occurs as the water changes to
         steam. The water-steam mixture remains at constant temperature,
         100C even though energy is being added until all the liquid has been
         converted to steam. The energy required to convert water to steam at
         100C is Q = mLv

      e) Above 100C: during this portion of the curve, no phase change
         occurs, so all the energy added is used to increase the temperature of
         the steam. The energy that must be added to raise the temperature of
         the steam is Q = mcsteam T.


EXERCISE 15
Suppose that the amount of heat removed when 3.0 kg of water freezes at 00C
were removed from ethyl alcohol at its freezing/melting point of -1140C. How
many kilograms of ethyl alcohol would freeze?

SOLUTION
We want the same amount of heat removed from the water as from the ethyl
alcohol, i.e., Qwater = Qalcohol
or (mLf)water = (mLf)alcohol




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                     Lf water              33.5 10 4 J/kg
m alcohol  m water                3.0 kg                                  9.3 kg
                    L f alcohol            10.8  10 4 J/kg

EXERCISE 16
A woman finds the front windshield of her car covered with ice at -120C. The ice
has a thickness of 4.5 x 10-4 m, and the windshield has an area of 1.25 m 2. The
density of ice is 917 kg/m3. How much heat is required to melt the ice?

SOLUTION
The heat required is Q = mLf + cmT, where m = V.
Thus, Q = VLf + cVT
                          –4                    4           3
Q =(917)(4.50  10             )(1.25){33.5  10 +[2.00  10 ](12.0}
    1.85 10 5 J

EXERCISE 17
Find the mass of water that vaporizes when 2.1 kg of mercury at 205 C is added
to 0.11 kg of water at 80.0C

SOLUTION
From the conservation of energy, the heat lost by the mercury is equal to the
heat gained by the water. As the mercury loses heat, its temperature decreases;
as the water gains heat, its temperature rises to its boiling point. Any remaining
heat gained by the water will then be used to vaporize the water.
According to Equation, the heat lost by the mercury is Qmercury  (cmT )mercury .
The heat required to vaporize the water is, from Equation, Qvap  (m vap L v )water .
Thus, the total amount of heat gained by the water is
                     Qwater  (cmT )water  (mvap Lv ) water .
                                 Qlost by  Qgained by
                                    mercury           water

                   (cmT ) mercury  (cmT ) water ( mvap Lv ) water
          where Tmercury  (205 C  100.0 C) and T water  (100.0 C  80.0 C)

The specific heats of mercury and water are given in Table, and the latent heat of
vaporization of water is given in Table. Solving for the mass of the water that
vaporizes gives
                                   T
           c m e r cur y m e r cur y m e r cur y c wa te rmwa te rTwa te r
                       m
m vap 
                                    ( Lv ) wa te r

           [139 J/(kg C)](2.10 kg)(105 C)  [4186 J/(kg C)](0.110 kg)(20.0 C )
      
                                                     22.6 10 5 J/kg

       9.49  10 –3 kg

EXERCISE 18
A cube of ice is taken from the freezer at 8.5º C and placed in a 95-g aluminum
calorimeter filled with 310 g of water at room temperature of 20.0°C. The final
situation is observed to be all water at 17.0°C. What was the mass of the ice
cube?


DEPARTMENT OF ENGINEERING                                                                  53
ESM 1224, SEMESTER 3, 2006/2007                                                                                CHAPTER 3



SOLUTION
The heat lost by the aluminum and 310 g of liquid water must be equal to the
heat gained by the ice in warming in the solid state, melting, and warming in the
liquid state.
                                                                                               
 mAl cAl Ti Al  Teq  mH2 O cH2 O Ti H2 O  Teq  mice  cice Tmelt  Ti ice   Lfusion  cH2 O Teq  Tmelt 
                                                                                                               
         0.095 kg 900     J kg  Cº 3.0 Cº   0.31 kg 4186 J kg  Cº 3.0 Cº 
mice                                                                                       9.90  10 3 kg
             2100 J kg  Cº 8.5Cº   3.3  10 5 J kg  4186 J kg  Cº 17 Cº 
                                                                                    

EXERCISE 19
A 0.2 kg piece of aluminum that has a temperature of -155C is added to 1.5 kg
of water that has a temperature of 3.0C. At equilibrium the temperature is 0.0C.
Ignoring the container and assuming that the heat exchanged with the
surroundings is negligible, determine the mass of water that has been frozen into
ice.

SOLUTION
Using the energy-conservation principle
     cAluminum mAluminum TAluminum  cWater mWater TWater                                mIce Lf, Water
                 Heat gained by aluminum                 Heat lost by water       Heat lost by water that freezes

Solving for mIce, taking values for the specific heat capacities from Table, and
taking the latent heat for water from Table , we find that
             cAluminum mAluminum TAluminum  cWater mWater TWater
mIce 
                                           Lf, Water


          9.00 102 J/  kg  C    0.200 kg  0.0 C   155 C  
                                                                        
                                  
                                 33.5 10 J/kg
                                            4



                           4186J/  kg  C   1.5 kg  3.0 C  0.0 C 
                                                                            0.027 kg
                                           33.5 104 J/kg

EXERCISE 20
An unknown material has a normal melting/freezing point of -25C and the liquid
phase has specific heat capacity of 160 J/kgC. One –tenth of a kilogram of the
solid at -25C is put into a 0.15kg aluminum calorimeter cup that contains 0.1kg
of glycerin. The temperature of the cup and the glycerin is initially 27C. All the
unknown material melts, and the final temperature at equilibrium is 20.0C. The
calorimeter neither losses energy nor gains energy from the external
environment. What is the latent heat of fusion of the unknown material?

SOLUTION
The system is comprised of the unknown material, the glycerin, and the
aluminum calorimeter. From the principle of energy conservation, the heat
gained by the unknown material is equal to the heat lost by the glycerin and the
calorimeter. The heat gained by the unknown material is used to melt the
material and then raise its temperature from the initial value of –25.0 °C to the
final equilibrium temperature of Teq  20.0 C .




DEPARTMENT OF ENGINEERING                                                                                           54
ESM 1224, SEMESTER 3, 2006/2007                                                  CHAPTER 3



                               Qgained by  Qlost by         Qlost by
                                 unknow n          gly cerine    calorim eter

                      m u Lf  c u mu Tu      c gl m gl Tgl  c alm al Tal
Taking values for the specific heat capacities of glycerin and aluminum from
Table, we have
(0.10 kg)Lf  [160 J/(kg C)](0.10 kg)(45.0 C)  [2410 J/(kg C)](0.100 kg)(7.0 C )
                                                      [9.0  10 2 J/(kg C)](0.150 kg)(7.0 C )

                           Lf  1.9  10 4 J/kg
Solving for   Lf yields,




DEPARTMENT OF ENGINEERING                                                            55

								
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