SPECIFIC HEAT CAPACITY AND CALORIMETRY



CHAPTER 17 NOTES – THERMOCHEMISTRY

SPECIFIC HEAT CAPACITY

1. Units of heat:

Joule: metric unit for heat/energy

calorie: amount of energy needed to change 1 gram of water 1°C

kilocalorie (Calorie or food calorie)

2. Specific Heat Capacity how much energy it takes to change 1 gram of something’s temp. by 1°C

low specific heat capacity: it takes less energy to raise the temp (ex: metal)

high specific heat capacity: it takes more energy to raise the temp (ex: water)

3. Specific Heat Capacity – think about it!

1. Which substance below would have the highest specific heat capacity? Which would have the lowest?

a. sand, water, air water = highest, sand = lowest

b. iron, granite, glass glass = highest, iron = lowest

2. Look at the table of specific heat capacities below.

|Substance |Specific Heat Capacity |

|aluminum |0.897 J/g°C |

|iron |0.454 J/g°C |

|lead |0.130 J/g°C |

Which substance heats up the quickest. __lead__________________________

If each of the substances above absorbs 1000J of heat, which would become the hottest? Why?

Lead – it takes the least amount of energy to raise its temperature

4. Practice specific heat problems:

a. The specific heat capacity of aluminum is 0.897 J/g°C. Determine the amount of heat released when 10.5g of aluminum cools from 240°C to 25°C.

H = m x (sh) x ΔT

= 10.5g (0.897 J/g°C) (215) = 2024.98 J

b. The specific heat capacity of aluminum is 0.897 J/g°C. What mass of aluminum can be heated from 33°C to 99°C using 450J of heat?

H = m x (sh) x ΔT

450 J = m (0.897 J/g°C) (66)

7.6g = m

Calorimetry: heat gained by water = heat lost by process

m(sh)(ΔT) = m(sh)(ΔT)

CALORIMETER

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1. 55.0g of a “mystery metal” at 93°C is placed in a calorimeter containing 100.0g of water at 25°C. The metal sits in the water until the temperature levels off at 29°C. At this point, both the metal and the water are at 29°C. The specific heat capacity of water is 4.18 J/g°C. Find the specific heat capacity of the metal.

| |METAL |WATER |

|mass |55g |100 g |

|specific heat capacity |? |4.18 J/g°C |

|Tinitial |29 °C |29 °C |

|Tfinal |93 °C |25 °C |

|ΔT |64 °C |4 °C |

a. Use H = m x (sh) x ΔT to find the amount of heat gained by the water.

H = 100(4.18)(4)

H = 1672 J

b. How much heat was lost by the metal?

1672 J (same amount of energy gained by the water!!)

c. Use H = m x (sh) x ΔT to find the specific heat capacity of the metal.

1672 = 55(sh)(64)

0.475 J/g°C = sh

2. 84.0g of a metal are heated to 112ºC, and then placed in a coffee cup calorimeter containing 60.0g of water at 32ºC. The final temperature in the calorimeter is 42ºC. What is the specific heat of the metal?

| |METAL |WATER |

|mass |84.0 g |60.0g |

|specific heat capacity | |4.18 J/g°C |

|Tinitial |112ºC |32ºC |

|Tfinal |42ºC |42ºC |

|ΔT |70 ºC |10 ºC |

m(sh)(ΔT) = m(sh)(ΔT)

84.0g(sh)(70 ºC) = 60.0g(4.18 J/g°C)(10 ºC)

sh = 0.43 J/g°C

__________________

3. The specific heat of aluminum is 0.902 J/gºC. 15.0g of aluminum are heated to 115ºC, and added to a calorimeter containing water at 25ºC. The final temperature in the calorimeter is 45ºC. What mass of water was in the calorimeter?

| |METAL |WATER |

|mass |15g |? |

|specific heat capacity |0.902 J/g°C |4.18 J/g°C |

|Tinitial |45 ºC |45 ºC |

|Tfinal |115 ºC |25 ºC |

|ΔT |70 ºC |20 ºC |

m(sh)(ΔT) = m(sh)(ΔT)

15(0.902)(70) = x(4.18)(20)

11.33g = x

___________________

ENERGY AND PHASE CHANGES

Vocabulary:

Solid most organized state of matter (particles close together/ most attracted); of all states, solids have lowest potential energy

Liquid flow, less organized, some attraction between particles, have more potential energy than solids

Gas particles move freely, very little attraction between particles, highest potential energy of all states of matter

Potential Energy energy of position (stored energy) – energy you have because you “are”

Kinetic Energy energy of motion (temperature)

Heat of Fusion energy necessary to melt one gram of a substance at its melting point, energy given off when one gram of a substance freezes at its freezing point

Heat of Vaporization energy needed to boil one gram of a substance at its boiling point, energy released if a gram of a substance condenses

Specific Heat Capacity heat needed to raise the temperature of one gram of a substance by 1°C

REVIEW:

1. The specific heat capacity of water is 4.18J/g°C. How much heat is needed to raise the temperature or 150g of water from 17°C to 55°C?

H = m(sh)(ΔT)

H = 150 g (4.18 J/g°C) (38°C)

23826 J

Is this process endothermic or exothermic? __endothermic__________

What would be the sign of ΔH? _____+________

HEATING CURVE FOR WATER

Time

The data below are for water (H2O)

|Melting point |Boiling Point |Heat of Fusion |Heat of Vaporization |SH Capacity (solid) |SH Capacity |SH Capacity (vapor) |

| | | | | |(liquid) | |

| | | | | | | |

|0.0°C |100.0°C |334 J/g |2260 J/g |2.05 J/gºC |4.18 J/gºC |1.90 J/gºC |

1. Draw a heating curve for substance water, going from -20ºC to 125ºC on the axis below. Write in all formulas used to calculate heat.

b. Determine the amount of heat necessary to convert 15.0g of water at 100.0ºC to steam at 100.0ºC.

H = mHv

H = 15(2260)

c. Determine the amount of heat necessary to convert 15.0g of ice at 0.0ºC to liquid at 0.0º C.

H = mHf

H = 15(334)

d. Determine the amount of heat given off when 25g of water cools from 56.2ºC to 33.5ºC.

H = m(sh)(ΔT)

H = 25g(4.18 J/gºC)(22.7ºC)

e. Determine the amount of heat needed to raise the temperature of 19.6g of X from -15ºC to -2.0ºC.

H = m(sh)(ΔT)

H = 19.6g(2.05 J/gºC)(13ºC)

MORE HEAT PRACTICE

The data below refers to an unknown substance, X.

|Melting point |Boiling Point |Heat of Fusion |Heat of Vaporization |SH Capacity (solid) |SH |SH (vapor) |

| | | | | |(liquid) | |

| | | | | | | |

|32.0°C |112.0°C |425 kJ/g |695 kJ/g |2.3 J/gºC |5.9 J/gºC |1.1 J/gºC |

2. Draw a heating curve for substance X, going from 15ºC to 125ºC on the axis below. Write in all formulas used to calculate heat.

b. Determine the amount of heat released when 15.0g of gaseous X at 112.0ºC changes to liquid at 112.0ºC.

H = mHv

H = 15g (695 kJ/g)

c. Determine the amount of heat necessary to convert 15.0g of solid X at 32.0ºC to liquid at 32.0º C.

H = mHf

H = 15g (425 kJ/g)

d. Determine the mass of X that can be heated from 38°C to 102°C using 4500J of heat.

H = m(sh)(ΔT)

4500 J = m (5.9 J/gºC) (64 °C)

e. Find the final temperature if 150g of X, at 19°C, receives 650J of heat.

H = m(sh)(ΔT)

650 J = 150 (2.3 J/gºC) (Tf - 19 °C)

MULTI-STEP HEAT PROBLEMS

The data below refers to an unknown substance, X.

|Melting point |Boiling Point |Heat of Fusion |Heat of Vaporization |SH Capacity (solid) |SH |SH (vapor) |

| | | | | |(liquid) | |

| | | | | | | |

|14.0°C |86.0°C |150 J/g |550 J/g |4.3 J/gºC |5.2 J/gºC |1.1 J/gºC |

1. Draw a heating curve for substance X, going from 2ºC to 120ºC on the axis below. Write in all formulas used to calculate heat.

b. Determine the amount of heat necessary to change 30g of X from solid at 4°C to liquid at 14°C.

STEP 1: STEP 2:

TOTAL:

c. How much heat will be released when 25g of X cools from 110°C to 50°C?

STEP 1: STEP 2:

STEP 3:

TOTAL:

THERMODYNAMICS: ENERGY & ENTROPY IN CHEMICAL REACTIONS

Enthalpy of a Reaction (ΔH): change in heat (Hproducts – Hreactants)

ΔH = “+” = endothermic = H absorbed

ΔH = “-“ = exothermic = H released

Entropy: (ΔS) – change in order/how disorganized a substance is

ΔS = “+” = system is more disorganized

ΔS = “-“ = system is less disorganized

Which process(es) below would have a positive change in entropy (ΔS)?

a. cleaning your room

- ΔS

b. a bottle breaking

+ ΔS

c. leaves falling off of a tree

+ ΔS

d. water freezing

- ΔS

States of matter, solutions and entropy:

Solid Most ordered

Liquid

Aqueous (solution)

Gas Less ordered

Which reactions below would have a positive change in entropy (ΔS)?

a. Al2(CO3)3(s) ⋄ Al2O3(s) + 3CO2(g)

Positive ΔS

b. KCl(aq) + Br2(l) ⋄ KBr(aq) + Cl2(g)

Positive ΔS

c. FeCl2(aq) + H2SO4(aq) ⋄ FeSO4(s) + HCl(aq)

Negative ΔS

How can you tell if a reaction will be spontaneous? (able to occur?)

|ΔS change in entropy |ΔH change in heat |Spontaneous??? |

|+ more disorder | exothermic releases heat |Yes – it can occur at any temp. |

|+ more disorder | + endothermic |Spontaneous at high temperature |

| |requires heat | |

|- less disorder/more order | - exothermic |Spontaneous at low temperature |

| |releases heat | |

|- less disorder/more order | + endothermic |No – it cannot occur |

| |requires heat | |

1. Determine whether each reaction will have a positive or negative change in entropy (ΔS).

2. Label each reaction as endothermic or exothermic.

Determine which reaction(s) below will naturally occur (be spontaneous)? Which will require added energy? Which will only occur naturally at low temperatures?

a. Na2CO3 (s) ⋄ Na2O (s) + CO2 (g) ΔS = __+___ ΔH = 321kJ

Spontaneous? ___@ high temp__

b. Mg (s) + 2HCl (aq) ⋄ MgCl2 (aq) + H2 (g) ΔS = _+___ ΔH = - 457.6kJ

Spontaneous? _ yes - always_____

c. N2 (g) + 3H2 (g) ⋄ 2NH3 (g) ΔS = _-____ ΔH= -91.8kJ

Spontaneous? __@ low temp____

d. 2KCl (s) + 3O2 (g) ⋄ 2KClO3 (s) ΔS =_-____ ΔH = 90.0kJ

Spontaneous? ___never______

HOMEWORK: CALCULATING ΔH FOR TEMPERATURE CHANGES USING SPECIFIC HEAT CAPACITY

1. Define the terms below:

specific heat capacity:

enthalpy:

2. Read the paragraph below. Then look over the sample problems.

The amount of heat gained or lost during a temperature change depends on three factors:

the amount of the substance heating or cooling, the identity of the substance, and how much the temperature changes. Those factors can be put together in the formula:

o ΔH is the amount of heat gained or lost

o m is the mass of the substance

o sh is the specific heat capacity (which is unique to each substance)

o ΔT is the temperature change (which is calculated by subtracting the initial temperature from the final temperature)

SAMPLE 1: Determine the amount of heat needed to raise the temperature of 15.0g of lead from 22°C to 68°C. The specific heat capacity of lead is 0.13 J/g°C.

Formula: ΔH = m x sh x ΔT

Substitution: ΔH = (15.0g) x (0.13J/g°C) x (46°C) Note: ΔT = 68°C - 22°C = 46°C

Answer: ΔH = 89.7 J

SAMPLE 2: What mass of water can be heated from 15°C to 75°C using 6500J of heat? The specific heat of water is 4.18 J/g°C.

Formula: ΔH = m x sh x ΔT

Substitution: 6500J = m x (4.18 J/g°C) x (60°C) Note: ΔT = 75°C - 15°C = 60°C

Calculator Math: m = 6500J ÷ 4.18 J/g°C ÷ 60°C

Answer: 25.9g water

3. Complete the problems on the back of this page.

For each problem below, show the formula, substitution and answer for complete credit. When you are done, check your answers. If they are not correct, go to the back of the room and ask your teacher for help. Your teacher will only help you if the formula and substitution are already written out for each problem.

a. What mass of lead can be heated from 23°C to 44°C using 125J of heat? The specific heat capacity of lead is 0.129 J/g°C.

ΔH = m x sh x ΔT

125 J = m (0.129 J/g°C) (21°C)

b. When 32.0g of a substance cools from 85°C to room temperature (25°C), 2400J of heat are released. Find the specific heat capacity of the substance.

ΔH = m x sh x ΔT

2400 J = 32 g (sh) (60°C)

c. A 16-g piece of iron absorbs 1090 joules of heat energy, and its temperature changes from 25ºC to 175ºC. Calculate the heat capacity of iron.

ΔH = m x sh x ΔT

1090 J = 16 g (sh) (150°C)

d. How many joules of heat are needed to raise the temperature of 10.0 g of aluminum from 22ºC to 55ºC, if the specific heat of aluminum is 0.90 J/gºC?

ΔH = m x sh x ΔT

ΔH = 10 g (0.90 J/g°C) (33°C)

e. To what temperature will a 50.0 g piece of granite raise if it absorbs 5275 joules of heat and its heat capacity is 0.50 J/gºC? The initial temperature of the granite is 20.0ºC.

ΔH = m x sh x ΔT

5275 J = 50 g (0.50 J/g°C) (Tf - 20°C)

f. Calculate the heat capacity of a piece of silver if 1500 g of the silver absorbs 6.80kJ of heat, and its temperature changes from 32ºC to 57ºC.

ΔH = m x sh x ΔT

6800 J = 1500 g (sh) (25°C)

ANSWERS: a. 46.1g Pb b. 1.25 J/g°C c. 0.454 J/g°C d. 297 J e. 231°C f. 0.181 J/g°C

HOMEWORK: CALORIMETRY PRACTICE

| |METAL |WATER |

|mass | | |

|specific heat capacity | |4.18 J/g°C |

|Tinitial | | |

|Tfinal | | |

|ΔT | | |

1. A piece of metal with a mass of 38.0 g and a temperature of 115°C is placed into 75.0 mL of water at 15°C. The final temperature of the system is 20°C. The density of water is 1.0 g/mL. What is the specific heat of the metal?

38g (sh) (95°C) = 75 g (4.18 J/g°C) (5°C)

sh = 0.43 J/g°C

2. A piece of gold (C= 0.129 J/g°C) is heated to 100°C and placed into 60 mL of water at 20.5°C. The final temperature of the system is 22°C. What was the mass of the metal?

m ( 0.129 J/g°C) (78°C) = (60g) (4.18 J/g°C) (1.5°C)

m = 37.4 g

3. 25.0 g of copper (C=.389 J/g°C) at 95°C is placed into water at 25°C. The final temperature of the system is 28.5°C. What was the mass of the water?

25g (0.389 J/g°C) (66.5°C) = m (4.18 J/g°C) (3.5°C)

44.2 g = m

4. A piece of aluminum (C=0.900 J/g°C) with a mass of 55.0 g is heated and placed into 45.0 mL of water at 18°C. The final temperature of the system is 22°C. What was the starting temperature of the metal?

55g (0.9 J/g°C) (ΔT) = 45g (4.18 J/g°C) (4°C)

ΔT = 15.2

15.2 + 22 = 37.2°C

HOMEWORK: HEAT PRACTICE

|Melting point |Boiling Point |Heat of Fusion |Heat of Vaporization |SH Capacity (solid) |SH Capacity |SH Capacity (vapor) |

| | | | | |(liquid) | |

| | | | | | | |

|-12.°C |76.0°C |65 kJ/g |88 kJ/g |5.2 J/gºC |2.1 J/gºC |1.7 J/gºC |

1. Draw a heating curve for substance X, going from -20ºC to 125ºC on the axis below. Write in all formulas used to calculate heat.

b. Determine the amount of heat released when 24.0g of liquid X at -12.0ºC changes to solid at -12.0ºC.

H = mHf

H = 24g (65 kJ/g)

c. What mass of X can be converted from liquid to vapor at 76.0ºC using 425kJ of heat?.

H = mHv

425kJ = m(88kJ)

d. Determine the amount of heat given off when 25g of X cools from 96.2ºC to 83.1ºC.

H = m(sh)(ΔT)

H = 25(1.7)(13.1)

e. Find the final temperature if 17g of X at 115°C absorbs 725J of heat.

H = m(sh)(ΔT)

725 = 17(1.7)(Tf-115)

HOMEWORK: ENTROPY AND SPONTANEITY OF CHEMICAL REACTIONS

1. Determine whether ΔS will be positive (disorder increases) or negative (disorder decreases) for each process below:

a. burning paper _____+________________

b. organizing your binder _____-________________

c. water freezing _____-________________

d. dry ice sublimating _____+________________

2. Determine the sign of ΔS for each reaction below. Then determine under which conditions (if at all) the reactions will be spontaneous.

a. CO2 (g) → C(graphite) + O2 (g) ΔH = +393.5 kJ/mol

ΔS = ___-_____ Spontaneous? ___ no ________

b. 2HCl(g) → H2 (g) + Cl2 (g) ΔH = +184.6 kJ/mol

ΔS = _________ Spontaneous?_____________

c. CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l) ΔH = -890.4 kJ/mol

ΔS = ___-_____ Spontaneous? __@ low temps___

d. 2 Na(s) + 2 H2O(l) ⋄ 2 NaOH (aq) + H2(g) ΔH = -446 kJ/mol

ΔS = ___+_____ Spontaneous?___ yes ______

3. A student tries to dissolve a salt (ionic compound) in water. He is having a hard time, so he decides to use the bunsen burner to heat up the water as the salt dissolves. It works! Determine the signs for ΔH and ΔS for this process.

ΔH __ positive ____________ ΔS ____ positive ___________

-----------------------

T

E

M

P

E

R

A

T

U

R

E

(ºC)

H = 33,900 J

ΔH = __+___

H = 5010 J

ΔH = __+___

H = 2372.15 J

ΔH = _-____

H = 522.34 J

ΔH = ___-__

H = 10425 J

ΔH = _-____

H = 6375 J

ΔH = __+___

m = 11.9 g

ΔH = __+___

Tf = 20.88 °C

ΔH = __+___

ΔH = m x sh x ΔT

H = 1560 J

ΔH = __-____

m = 4.83g

ΔH = __+___

H = 556.75 kJ

ΔH = __-___

Tf = 140.09°C

ΔH = __+___

Liquid ( Vapor (gas) (boiling)

Increase PE

H = mHvaporization

ΔS is +

ΔS is -

Solid ( Liquid (melting)

Increase PE (pulling attractions of particles apart requires energy)

H = mHfusion

Vapor (gas)

Increase KE

H = m(sh)(ΔT)

Liquid

Increase KE

H = m(sh)(ΔT)

Solid

Increase KE (temp change)

H = m(sh)(ΔT)

125°C

100°C

0°C

-10°C

H = mHv

H = mHf

H = m(sh)(ΔT)

H = m(sh)(ΔT)

H = m(sh)(ΔT)

H = mHv

H = mHf

H = m(sh)(ΔT)

H = m(sh)(ΔT)

H = m(sh)(ΔT)

125°C

112°C

32 °C

0°C

H = mHv

H = mHf

H = m(sh)(ΔT)

H = m(sh)(ΔT)

H = m(sh)(ΔT)

100°C

76°C

-12 °C

-25°C

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