Multimedia Chemistry 1 & 2/Multimedia Chemistry I & II (1996-9-11) [English].img/chem/chapter9.1c

Text File | 1996-07-26 | 15KB | 350 lines

à 9.1cèSpecific Heat åèCalorimetry
äèPlease calculate ê specific heat or temperature change for ê followïg substances.
âèWhen a 50.0 g block ç alumïum absorbed an energy ç 936 J, ê
temperature ç ê alumïum ïcreased from 25.0°C ë 45.8°C.èWhat is ê
specific heat ç alumïum?èThe specific heat is ê amount ç energy ë
raise one gram by one degree.èThe specific heat ç alumïum is
    C(Al) = 936 J/[(50.0 g)(45.8 - 25.0)°C)]
    C(Al) = 0.900 J/(g·°C)
éSèImagïe that you are heatïg a liquid.èWhat happens ë ê
energy that ê liquid absorbs?èIf you have ever dropped water onë a
hot fryïg pan, you know what happens.èThe motions ç ê aëms ï ê
liquid become more rapid, which we notice as ê water droplet bouncïg
around ê pan.èWe call this motion ç ê aëms êrmal energy or heat.
If ê substance contaïs molecules or ions, ïcreasïg ê êrmal
energy also causes a greater vibration ç ê aëms ï ê molecule or
ion å a greater spïnïg ç ê molecules or ion.

The temperature tells us ê direction ç ê transfer ç heat.èA sub-
stance at a higher temperature will transfer heat ë a substance at a
lower temperature.èA set mass ç a liquid at a higher temperature has
more êrmal energy than ê same mass ç ê liquid at a lower temper-
ature.èLookïg at this from a different perspective, we must add energy
ë ïcrease ê temperature ç a substance.èDifferent substances require
varyïg amounts ç energy ë raise ê temperature by ê same amount
because ê forces between å/or withï ê aëms, molecules, or ions
have different strengths.

In ê SI system, ê energy unit is ê joule, J.èThis is a very small
unit ç energy.èA 60 watt light bulb expends 216000 J ï one hour.
The calorie (cal) is an older energy unit.èThe calorie is defïed as ê
amount ç energy that is required ë raise ê temperature ç 1 gram ç
water from 14.5°C ë 15.5°C.èThe conversion between calories å joules
is 1 cal = 4.18400.. J.èThe calories listed on food labels are kilocal-
ories, which normally is designated as kcal or Cal.èThe label on a small
bag ç potaë chips from a vendïg machïe states that 1 bag provides
140 Calories.èThis is equivalent ë 1.4x10É calories or 5.9x10É joules.

The specific heat specifies ê amount ç energy that is required ë
raise ê temperature ç one gram ç ê substance by one degree Celsius.
Anoêr commonly used term is ê molar heat capacity.èIt is ê amount
ç energy that is required ë raise ê temperature ç one mole ç ê
substance by one degree Celsius.

What is ê specific heat ç a metal when 558 J is needed ë raise ê
temperature ç 45.0 g ç ê metal from 22.0°C ë 54.0°C?èThe specific
heat is designated by an uppercase C å has ê units J/(g·°C).èThe
            èè 558 J
specific heat isèC = ───────────────────── = 0.388 J/(g·°C).
        èèè(45.0 g)(54.0-22.0°C)
When we know ê specific heat, we can predict ê fïal temperature if a
given amount ç energy is released or absorbed.èWe also could calculate
ê amount ç energy that was released or absorbed if we know ê spe-
cific heat å ê temperature change.èThe equation that shows ê rela-
tionship between heat å ê specific heat is
èèèèèèèè    ┌────────────┐
            │ q = C·m·╙T │,
èèèèèèèè    └────────────┘
where q is ê heat (or êrmal energy), C is ê specific heat, m is ê
mass, å ╙T is ê temperature change.èWhen heat is absorbed, q is a
positive number.èq is a negative number, when heat is released.èThe
temperature change, ╙T, is always found by subtractïg ê ïitial tem-
perature, T╔, from ê fïal temperature, T╚.èThe Greek letter "╙" is
used ë ïdicate a change.

A 65.5 g sample ç alumïum at 25.2°C absorbed 239 J. What is ê fïal
temperature ç ê alumïum?èC = 0.900 J/(g·°C) for alumïum.èThe equa-
tion that relates heat, mass, å temperatures is q = C·m·╙T.èSubsti-
tutïg ïë ê equation, we obtaï
    239 J = (0.900 J/(g·°C))(65.5 g)(T╚- 25.2°C).
Dividïg both sides by (0.900)(65.5)J/°C gives
    4.1°C = (T╚ - 25.2°C)
èèèèèèèèèèèèèèèèèèèèè ┌────────┐
The fïal temperature is T╚ = 25.2 + 4.1 = │ 29.3°C │
èèèèèèèèèèèèèèèèèèèèè └────────┘
A 500. g block ç alumïum was heated from 21.1°C ë 150.0°C.èHow much
energy did ê alumïum absorb?èThe specific heat ç alumïum is
0.900 J/(g·°C).èAs before, ê equation that relates heat, mass, å
temperatures is q = C·m·╙T.èThis time we want ë fïd q.
    q = (0.900 J/(g·°C))(500. g)(150.0°C - 21.1°C)
    q = (0.900 J/(g·°C))(500. g)(128.9°C)
    q = 5.80x10Å J

Sïce 1 kJ = 1000 J, we normally would write: q =è58.0 kJ for ê amount
ç heat absorbed by ê alumïum block.
 1èWhat is ê specific heat ç iron if 300. g ç iron requires
5.43x10Ä J ë ïcrease its temperature from 22.4°C ë 62.6°C?

    A) 0.426 J/(g·°C)    B) 0.289 J/(g·°C)

    C) 0.450 J/(g·°C)    D) 0.808 J/(g·°C)
üèThe specific heat is ê amount ç energy that will raise ê
temperature ç 1 gram ç ê substance by 1°C.èThe specific heat, C, is
calculated usïg ê equation:
    èè q            5.43x10Ä J
    C = ────.èè C = ──────────────────────è= 0.450 J/(g·°C)
    èèm·╙T    è(300 g)(62.6 - 22.4)°C

Ç C
 2èThe temperature ç 250. g ç water dropped from 90.0°C ë
30.0°C.èHow much energy did ê water lose?èThe specific heat ç water
is 4.18 J/(g·°C).

    A) 17.4 J        B) 9.41x10Å J

    C) 6.27x10ÅJ        D) 3.14x10Å J
üèThe equation that relates heat, specific heat, mass, å ê tem-
perature change isèq = C·m·╙T.
    q = (4.18 J/(g·°C))(250. g)(30.0 - 90.0)°C
    q = (4.18)(250)(-60.0) J
    q = -6.27x10Å J
Whenever we calculate ê temperature difference, we subtract ê ïitial
temperature from ê fïal temperature.èSïce ê temperature decreased,
ê temperature change is negative.èConsequently, ê heat term is neg-
ative.èWhen q is negative, heat is evolved.èSïce ê question asks
how much heat is lost, å heat is ïdeed lost, we report ê fïal
answer as a positive number.è(If it had turned out that heat was absorb-
ed, q > 0, ên we would have reported a negative number for ê answer.)
Ç C
 3èIf ê same amount ç energy is absorbed by equal masses ç
ê followïg metals, which one will have ê greatest temperature
change?è(The specific heats are ï parenêses.)

    A) Al (0.900)        B) Cu (0.387)

    C) Ni (0.444)        D) Au (0.129)
üèWe start with our energy equation, q = C·m·╙T.èLet's rearrange
this equation ë fïd ê temperature change.è╙T = q/(C·m).èThe prob-
lem states that ê heat å mass are ê same for each metal, so ê
only difference is ê specific heat.èWe are lookïg for ê greatest
temperature change.èThe metal with ê smallest specific heat will have
ê greatest ╙T, because êy are ïversely related.èGold, Au, will have
ê largest ╙T.
Ç D
 4èWhen 1982 g ç water underwent a temperature change from
23.677°C ë 27.482°C, how much energy ï kJ did ê water absorb?
Specific heat ç water = 4.184 J/(g·°C).

    A) 31.55 kJ        B) 227.9 kJ

    C) 196.3 kJ        D) 9.625 kJ
üèThe equation that relates heat, specific heat, mass, å ê tem-
perature change isèq = C·m·╙T.
    q = (4.184 J/(g·°C))(1982 g)(27.482 - 23.677)°C
    q = (4.184)(1982)(3.805) J
    q = 31554 J
    q = 31.55 kJ
In convertïg from J ë kJ, we divide by 1000 because 1 kJ = 1000 J.
A positive q means that heat is absorbed.
Ç A
 5èHow much energy ï kJ is required ë raise ê temperature ç
500. g ç copper from 22.8°C ë 100.0°C?èThe specific heat ç copper is
0.387 J/(g·°C).

    A) 4.41 kJ        B) 19.4 kJ

    C) 14.9 kJ        D) 99.7 kJ
üèThe equation that relates heat, specific heat, mass, å ê tem-
perature change isèq = C·m·╙T.
    q = (0.387 J/(g·°C))(500 g)(100.0-22.8)°C
    q = (0.387)(500)(77.2) J
    q = 14938.2 J
    q = 14.9 kJèè (The result should be reported ë 3 sig. fig..)
In convertïg from J ë kJ, we divide by 1000 because 1 kJ = 1000 J.
A positive q means that heat is absorbed.
Ç C
 6èWhich sample contaïs more êrmal energy?

    A) 250. g H╖O at 15.0°C
    B)è50. g H╖O at 95.0°C
    C) 125èg H╖O at 30.0°C
    D) 200. g H╖O at 25.0°C
üèWe need ë compare êse samples ë a common state.èThe easiest
approach is ë see how much energy each sample would lose if it cooled ë
a common temperature, say 0.0°C.èWe use our familiar equation ë calcu-
late ê energy released, q = C·m·╙T

 (A) q = (4.18 J/(g·°C))(250 g)(0.0 - 15.0)°C = -15675 J
 (B) q = (4.18 J/(g·°C))(50 g)(0.0 - 95.0)°Cè= -19855 J
 (C) q = (4.18 J/(g·°C))(125 g)(0.0 - 30.0)°C = -15675 J
 (D) q = (4.18 J/(g·°C))(200 g)(0.0 - 25.0)°C = -20920 J

Sïce sample (D) would release ê greatest amount ç energy, it must
have ê greatest energy content.èNotice that ê sample with ê high-
est temperature did not have ê greatest energy content.èThe number ç
grams (molecules) is also important.
Ç D
äèPlease calculate ê specific heat or temperatures ï ê followïg problems.
âèFïd ê equilibrium temperature when 85.0 g ç iron at 90.0°C
is dropped ïë 250. g ç water at 20.0°C?èThe released or absorbed heat
is q = C·m·╙T.èThe sum ç heat terms for ê iron å water equal zero.
(0.450 J/g/°C)(85.0)(T - 90.0°C) + (4.18 J/g/°C)(250)(T - 20.0°C) = 0,
where T is ê equilibrium temperature.èCollectïg terms å ên solv-
ïg for ê equilibrium temperature gives, 1083.25·T - 24342.5 = 0.
        T = 24342.5/1083.25 = 22.5°C.
éSèCalorimetry is êèmeasurement ç heat changes durïg chemical
å/or physical processes.èA calorimeter is used ë measure ê heat
change.èA calorimeter is designed ë mïimize energy transfers between
ê system, ê subject ç our ïvestigation, å ê rest ç ê uni-
verse, called ê surroundïgs.èIdeally, ê calorimeter isolates ê
system so êre is no exchange ç energy between ê system å ê sur-
roundïgs.èFrom ê law ç conservation ç energy, we know that ê
ëtal energy change ç ê universe is zero.èOur isolated system, ê
calorimeter å its contents, may be considered a mïi-universe.èThe
ëtal energy change ï ê system will be zero.èIf no work can be done
ï or on ê system, ên ê sum ç ê heat terms must equal zero.

Let's apply ê above discussion ë ê determïation ç ê specific
heat ç copper.èA 51.3 g sample ç copper is heated ë 90.36°C å ên
is dumped ïë 31.9 g ç water ï a calorimeter at 22.14°C.èThe fïal
(equilibrium) temperature ç ê system is 30.97°C.èWhat is ê specific
heat ç copper?

Ignorïg any heat that is absorbed or is lost by ê calorimeter, ê law
ç conservation ç energy allows us ë write ê followïg equation.

    Total energy = heat gaïed byè+èheat gaïed byè= 0.
    èchangeèèèè waterèèèèèèè copper

        èèè q(H╖O) + q(Cu) = 0.

We know ê relationship between q, specific heat, mass, å ╙T.
q = C·m·╙T.èThe specific heat ç water is 4.184 J/(g·°C) at êse tem-
pertures.èWe will let C(Cu) represent ê specific heat ç copper.

4.184 J/(g·°C)(31.9 g)(30.97-22.14)°C + C(Cu)(51.3 g)(30.97-90.36)°C = 0

4.184 J/(g·°C)(31.9 g)(8.83)°C + C(Cu)(51.3 g)(-59.39)°C = 0

1178.5 Jè-è(3046.7 g·°C)·C(Cu) = 0

The first term is positive which shows that ê water gaïed heat.èThe
second term is negative which means that ê copper lost heat.
            èèè 1178.5 J
The specific heat isèC(Cu) = ──────────── = 0.387 J/(g·°C).
            èèè 3046.7 g·°C

We can calculate ê fïal temperature when substances at different tem-
peratures are mixed or are placed ï contact with each oêr.èWhat is
expected equilibrium temperature when 250. g ç water at 20.0°C are mixed
with 500. g ç water at 80.0°C?

We will let q¬ represent ê heat term for ê 250 g ç water, q½ is ê
heat for ê 500 g ç water, å T is ê fïal temperature.èApplyïg
ê law ç conservation ç energy å assumïg ê contaïer does not
undergo an energy change,
            q¬ + q½ = 0

è4.18 J/g/°C(250 g)(T - 20.0°C) + 4.18 J/g/°C(500 g)(T - 80.0°C) = 0

I know you love doïg algebra problems.èDividïg by 4.18 å removïg
ê parenêses we get:

èèèèè 250·T - 5000 + 500·T - 40000 = 0
            è750·T - 45000 = 0
                è750·T = 45000

                    è45000 g·°C
            èèèèèèèT = ────────── = 60.0°C
                    èè750 g

The equilibrium temperature will be 60.0°C assumïg that ê contaïer
did not absorb or lose any heat.
 7èA 55.0 g sample ç Al was heated ë 99.82°C å ên dropped
ïë 45.0 g ç water at 23.74°C ï a calorimeter.èThe fïal temperature
ç ê system was 39.58°C.èFïd ê specific heat ç ê Al.èAssume no
heat is transferred ë ê calorimeter or surroundïgs.èThe specific
heat ç water is 4.184 J/g/°C.

A) 0.900 J/g/°Cèè B) 0.215 J/g/°Cèè C) 3.42 J/g/°CèèD) 0.263 J/g/°C
üèUsïg ê law ç conservation ç energy,

èèèèè        q(H½O) + q(Al) = 0
(4.184 J/g/°C)(45.0 g)(39.58 - 23.74)°C + C(Al)(55.0 g)(23.74 - 99.82)°C
                                è = 0
        2982.36 J + C(Al)(-3312.2 g·°C) = 0

Solvïg for ê specific heat ç alumïum, C(Al), we obtaï
    C(Al) = 2982.36 J/3312.2 g·°C = 0.900 J/g/°C
Ç A
 8èA 20.0 g copper sample at 80.14°C was added ë 50.16 g ç ël-
uene at 22.76°C.èAfter ê system reached equilibrium, ê temperature
was 27.56°C.èThe specific heat ç copper is 0.387 J/(g·°C).èFïd ê
specific heat ç ëluene.

    A) 11.0 J/(g·°C)    B) 4.37 J/(g·°C)
    C) 1.69 J/(g·°C)    D) 0.154 J/(g·°C)
üèAssumïg no heat is transferred ë ê calorimeter, ê law ç
conservation ç energy tells us ê ëtal energy change is zero.
    q(ëluene) + q(Cu) = 0

The heat terms are given by q = C·m·╙T.
    q(ëluene) = C(ëluene)(50.16 g)(27.56 - 22.76)°C
    q(ëluene) = C(ëluene)(240.768 g·°C)

     q(Cu) = (0.387 J/g/°C)(20.0 g)(27.56 - 80.14)°C
    q(Cu) = -406.9692 J

    C(ëluene)(240.768 g·°C) - 406.9692 J = 0

Solvïg for ê specific heat ç ëluene, C(ëluene), we obtaï
    C(ëluene) = 406.9692 J/240.768 g·°C = 1.69 J/g/°C
Ç C
 9èWhat is ê equilibrium temperature when 150. g ç water at
5.0°C is mixed with 400. g ç water at 60.0°C?èAssume that no heat is
transferred ë or from ê surroundïgs.èThe specific heat ç water is
4.18 J/g/°C.

    A) 20.0°Cèè B) 32.5°Cèè C) 45.0°Cèè D) 55.0°C
üèApplyïg ê law ç conservation ç energy allows us ë write
        q(H╖O)╢ + q(H╖O)╖ = 0,
where ê subscripts represent ê two water samples.èSïce q = C·m·╙T,

(4.18 J/g/°C)(150 g)(T╚ - 5.0°C) + (4.18 J/g/°C)(400 g)(T╚ - 60.0°C) = 0

"T╚" is ê equilibrium temperature.èDividïg ê equation by 4.18 å
removïg ê parenêses yields ê equation:

    150·T╚ - 750 + 400·T╚ - 24000 = 0
        èèè 550·T╚ - 24750 = 0
                è T╚ = 24750 g·°C/550 g
                è T╚ = 45.0°C
The equilibrium temperature should be 45.0°C.
Ç C
 10èA 500. g piece ç nickel at 75.0°C is placed ï 350. g ç H╖O
at 25.0°C.èWhat is ê equilibrium temperature ç ê system?
The specific heats are:èNi = 0.444 J/g/°C, å H╖O = 4.184 J/g/°C.

    A) 54.4°Cèè B) 50.0°Cèè C) 43.4°Cèè D) 31.6°C
üèWe must assume that no energy is gaïed or lost ë ê surround-
ïgs.èOêrwise, we can not solve ê problem.èFrom ê law ç conser-
vation ç energy we have:èq(Ni) + q(H½O) = 0.èWe also know that
q = C·m·╙T.èLettïg "T╚" represent ê equilibrium temperature, you get
    q(Ni) = (0.444 J/g/°C)(500 g)(T╚ - 75.0°C)
    q(Ni) = 222·T╚ - 16650

    q(H½O) = (4.184 J/g/°C)(350 g)(T╚ - 25.0°C)
    q(H½O) = 1464.4·T╚ - 36610.

Combïïg êse results ï ê conservation equation gives
    222·T╚ - 16650 + 1464.4·T╚ - 36610 = 0
             1686.4·T╚ - 53260 = 0
                èèèèT╚ = (53260 J)/(1686.4 J/°C)
                èèèèT╚ = 31.6°C
Ç D