Calorimetry

Heat Transfer

Al Al

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 20 gT = 20oC

20 g (40oC) 20 g (20oC) 30oC

Block “A” Block “B”Final

Temperature

Assume NO heat energy is “lost” to the surroundings from the system.

C30

g) 20 g (20C20g 20C40g 20 o

oo

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ?

Heat Transfer

AlAl

m = 20 gT = 40oC

SYSTEM

Surroundings

m = 10 gT = 20oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

Assume NO heat energy is “lost” to the surroundings from the system.

20 g (40oC) 10 g (20oC) 33.3oC

C3.33

g) 10 g (20C20g 10C40g 20 o

oo

What will be the final temperature of the system ?

a) 60oC b) 30oC c) 20oC d) ?

?

Heat Transfer

AlAl

m = 20 gT = 20oC

SYSTEM

Surroundings

m = 10 gT = 40oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

Assume NO heat energy is “lost” to the surroundings from the system.

20 g (40oC) 10 g (20oC) 33.3oC

C7.26

g) 10 g (20C40g 10C20g 20 o

oo

20 g (20oC) 10 g (40oC) 26.7oC

Heat Transfer

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

20 g (40oC) 20 g (20oC) 30.0oC

Block “A” Block “B”Final

Temperature

20 g (40oC) 10 g (20oC) 33.3oC

C46

g) 30 g (75C100g 30C25g 75 o

oo

20 g (20oC) 10 g (40oC) 26.7oC

AgH2O

Real Final Temperature = 26.6oC

Why?

We’ve been assuming ALL materialstransfer heat equally well.

Specific Heat

• Water and silver do not transfer heat equally well. Water has a specific heat Cp = 4.184 J/goC Silver has a specific heat Cp = 0.235 J/goC

• What does that mean? It requires 4.184 Joules of energy to heat 1 gram of water 1oC and only 0.235 Joules of energy to heat 1 gram of silver 1oC.

• Law of Conservation of Energy… In our situation (silver is “hot” and water is “cold”)… this means water heats up slowly and requires a lot of energy whereas silver will cool off quickly and not release much energy.

• Lets look at the math!

“loses” heat

Calorimetry

C26.6 x

320.8x 8550

7845 313.8x x 05.7 705

algebra. the solve and units Drop

C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0

equation. into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

Tfinal = 26.6oC

Calorimetry

C26.6 x

8550 320.8x

7845 313.8x x 05.7 705

algebra. the solve and units Drop

C25 -x g 75CgJ 184.4 C100 -x g 30CgJ 235.0

equation. into values Substitute

TTmC TTmC

TmC TmC

q q

o

oooo

ifpinitialfinalp

pp

OHAg 2

m = 75 gT = 25oC

SYSTEM

Surroundings

m = 30 gT = 100oC

AgH2O

240 g of water (initially at 20oC) are mixed with an unknown mass of iron (initially at 500oC). When thermal equilibrium is reached, the system has a temperature of 42oC. Find the mass of the iron.

Calorimetry Problems 2 question #5

FeT = 500oCmass = ? grams

T = 20oC

mass = 240 g LOSE heat = GAIN heat-

- [(Cp,Fe) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(0.4495 J/goC) (X g) (42oC - 500oC)] = (4.184 J/goC) (240 g) (42oC - 20oC)]

Drop Units: - [(0.4495) (X) (-458)] = (4.184) (240 g) (22)

205.9 X = 22091

X = 107.3 g Fe

A 97 g sample of gold at 785oC is dropped into 323 g of water, which has an initial temperature of 15oC. If gold has a specific heat of 0.129 J/goC, what is the final temperature of the mixture? Assume that the gold experiences no change in state of matter.

Calorimetry Problems 2 question #8

AuT = 785oCmass = 97 g

T = 15oC

mass = 323 g

LOSE heat = GAIN heat-

- [(Cp,Au) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(0.129 J/goC) (97 g) (Tf - 785oC)] = (4.184 J/goC) (323 g) (Tf - 15oC)] Drop Units:

- [(12.5) (Tf - 785oC)] = (1.35x 103) (Tf - 15oC)]

-12.5 Tf + 9.82 x 103 = 1.35 x 103 Tf - 2.02 x 104

3 x 104 = 1.36 x 103 Tf

Tf = 22.1oC

If 59 g of water at 13oC are mixed with 87 g of water at 72oC, find the final temperature of the system.

Calorimetry Problems 2 question #9

T = 13oC

mass = 59 g

LOSE heat = GAIN heat-

- [(Cp,H2O) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(4.184 J/goC) (59 g) (Tf - 13oC)] = (4.184 J/goC) (87 g) (Tf - 72oC)] Drop Units:

- [(246.8) (Tf - 13oC)] = (364.0) (Tf - 72oC)]

-246.8 Tf + 3208 = 364 Tf - 26208

29416 = 610.8 Tf

Tf = 48.2oC

T = 72oC

mass = 87 g

A 322 g sample of lead (specific heat = 0.138 J/goC) is placed into 264 g of water at 25oC.If the system's final temperature is 46oC, what was the initial temperature of the lead?

Calorimetry Problems 2 question #12

PbT = ? oCmass = 322 g

Ti = 25oC

mass = 264 g

LOSE heat = GAIN heat-

- [(Cp,Pb) (mass) (T)] = (Cp,H2O) (mass) (T)

- [(0.138 J/goC) (322 g) (46oC - Ti)] = (4.184 J/goC) (264 g) (46oC- 25oC)] Drop Units:

- [(44.44) (46oC - Ti)] = (1104.6) (21oC)]

- 2044 + 44.44 Ti = 23197

44.44 Ti = 25241

Ti = 568oC

Pb

Tf = 46oC

25 g of 116oC steam are bubbled into 0.2384 kg of water at 8oC. Find the final temperature of the system.

Calorimetry Problems 2 question #11

- [(Cp,H2O) (mass) (T)] + (Cv,H2O) (mass) + (Cp,H2O) (mass) (T) = [(Cp,H2O) (mass) (T)]

- [ - 816.8 - 56400 + 104.5Tf - 10450] = 997Tf - 7972

- [qA + qB + qC] = qD

qA = [(Cp,H2O) (mass) (T)]

qA = [(2.042 J/goC) (25 g) (100o - 116oC)]

qA = - 816.8 J

qB = (Cv,H2O) (mass)

qA = (2256 J/g) (25 g)

qA = - 56400 J

qC = [(Cp,H2O) (mass) (T)]

qC = [(4.184 J/goC) (25 g) (Tf - 100oC)]

qA = 104.5Tf - 10450

qD = (4.184 J/goC) (238.4 g) (Tf - 8oC)

qD = - 997Tf - 7972

- [qA + qB + qC] = qD

816.8 + 56400 - 104.5Tf + 10450 = 997Tf - 7972

67667 - 104.5Tf = 997Tf - 7979

75646 = 1102Tf

1102 1102

Tf = 68.6oC

Tem

pera

ture

(o C

)

40200

-20-40-60-80

-100

120100

8060

140

Time

H = mol x Hfus

H = mol x Hvap

Heat = mass x t x Cp, liquid

Heat = mass x t x Cp, gas

Heat = mass x t x Cp, solid

A

B

C

D

(1000 g = 1 kg)

238.4 g

A Coffee Cup Calorimeter

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 302

Thermometer

Styrofoamcover

Styrofoamcups

Stirrer

Thermometer

Glass stirrer

Cork stopper

Two Styrofoam ®cups nestedtogether containingreactants in solution

Bomb Calorimeter

thermometer

stirrer

full of water

ignition wire

steel “bomb”

sample

A Bomb Calorimeter