section 15-1 the nature of energy energy is the ability to do work or produce heat.energy...

Section 15-1

The Nature of Energy• Energy is the ability to do work or produce

heat.

• weightless, odorless, tasteless• Two forms of energy exist, potential and

kinetic.

• Potential energy is due to composition or position.

• Chemical potential energy is energy stored in a substance because of its composition.

• Kinetic energy is energy of motion.

Section 15-1

The Nature of Energy (cont.)

• The law of conservation of energy states that in any chemical reaction or physical process, energy can be converted from one form to another, but it is neither created nor destroyed.

• Heat is energy that is in the process of flowing (transferring) from a warmer object to a cooler object.

• q is used to symbolize heat.

• Essentially all chemical reactions and changes in physical state involve either:

• release of heat, or• absorption of heat

Endothermic and Exothermic Processes

In studying heat changes, think of defining these two parts:

•the system - the part of the universe on which you focus your attention

•the surroundings - includes everything else in the universe

•Together, the system and it’s surroundings constitute the universe


•Heat flowing into a system from it’s surroundings:

•defined as positive

•q has a positive value

•called endothermic

•system gains heat (gets warmer) as the surroundings cool down



•Heat flowing out of a system into it’s surroundings:

•defined as negative

•q has a negative value

•called exothermic

•system loses heat (gets cooler) as the surroundings heat up

Section 15-1

Measuring Heat

• A calorie is defined as the amount of energy required to raise the temperature of one gram of water one degree Celsius.

• Food is measured in Calories, or 1000 calories (kilocalorie).

• A joule is the SI unit of heat and energy, equivalent to 0.2390 calories.

• 1 calorie = 4.184 J or 1 J = 0.2390 calories

Section 15-1

Measuring Heat (cont.)

• Example:• A candy bar has 245 Calories. Convert this to calories

and then to Joules of energy.

Section 15-1

Specific Heat

• The specific heat of any substance is the amount of heat required to raise one gram of that substance one degree Celsius.

• Some objects require more heat than others to raise their temperature.

Section 15-1

Specific Heat (cont.)

• Calculating heat absorbed and released

– q = c × m × ΔT

– q = heat absorbed or released (in Joules)

– c = specific heat of substance

– m = mass of substance in grams

– ΔT = change in temperature in Celsius

• Examples:• How much heat does a 20.0 g ice cube absorb as its

temperature increases from (-27.0oC) to 0.0oC? Give

your answer in both joules and calories.

• q = c × m × ΔT

• Specific Heat of Ice = 2.03 J/goC

• 1 calorie = 4.184 J


• Example Cont. q = ? c = 2.03 J/goC m = 20.0 grams ΔT = FinalTemp(0.0oC) – InitialTemp (-27.0oC) = Change (27.0oC)

q = c × m × ΔT

q = (2.03 J/goC)(20.0g)(27.0oC)=


• Example 2: • A 5.00 gram sample of a metal is initially at 55.0 ºC.

When the metal is allowed to cool for a certain time, 98.8 Joules of energy are lost and the temperature decreases to 11.0º C. What is the specific heat of the metal? What metal is it?

• q = c × m × ΔTTo make the problem easier, solve for the unknown BEFORE you plug in the numbers.

• For Water during a phase change:– The Heat of Fusion (melting) is 334 j/g– The Heat of Solidification (freezing) is 334 j/g

• They are the same value (energy in or out)

– The Heat of Vaporization is 2260 j/g– The Heat of Condensation is 2260 j/g

• They are the same value (energy in or out)

Measuring Heat

• Example – Phase change

• Calculate the amount of energy needed to convert 55.0 grams of ice to all liquid water at its normal melting point.

• Using the same amount of water calculate the energy needed to completely vaporize the water at its normal boiling point.

• Why is there such a large difference in energy needed?

The solid temperature is rising from -20 to 0 oC (use q = mass x ΔT x C)

The solid is melting at 0o C; no temperature change (use q = mass x ΔHfus.)

The liquid temperature is rising from 0 to 100 oC (use q = mass x ΔT x C)

The liquid is boiling at 100o C; no temperature change (use q = mass x ΔHvap.)

The gas temperature is rising from 100 to 120 oC (use q = mass x ΔT x C)The Heat Curve for Water, going from -20 to 120 oC,

120

Section 15-2

Calorimetry

• Calorimetry - the measurement of the heat into or out of a system for chemical and physical processes.

• A calorimeter is an insulated device used for measuring the amount of heat absorbed or released in a chemical reaction or physical process.

• Based on the fact that the heat released = the heat absorbed

Section 15-2

Chemical Energy and the Universe (cont.)

• Chemists are interested in changes in energy during reactions.

• Enthalpy is the heat content of a system at constant pressure.

• Enthalpy (heat) of reaction is the change in enthalpy during a reaction symbolized as ΔHrxn.

ΔHrxn = Hfinal – Hinitial

ΔHrxn = Hproducts – Hreactants

Changes in enthalpy = H

q = H These terms will be used interchangeably in this textbook

Thus, q = H = m x C x T

H is negative for an exothermic reaction

H is positive for an endothermic reaction

Section 15-2

Chemical Energy and the Universe (cont.)

• The products are lower in energy than the reactants

• Thus, energy is released.

• ΔH = -395 kJ– The negative sign does not mean negative

energy, but instead that energy is lost.

Exothermic

• The products are higher in energy than the reactants

• Thus, energy is absorbed.

• ΔH = +176 kJ– The positive sign means energy is

absorbed

Endothermic

Chemistry Happens in

MOLES An equation that includes energy is

called a thermochemical equation CH4 + 2O2 CO2 + 2H2O + 802.2 kJ

• 1 mole of CH4 releases 802.2 kJ of energy.

• When you make 802.2 kJ you also make 2 moles of water and 1 mole of CO2

Thermochemical Equations

The heat of reaction is the heat change for the equation, exactly as written• The physical state of reactants and

products must also be given.

• Standard conditions (SC) for the reaction is (1 atm.) and 25 oC (different from STP)

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) + 802.2 kJ

If 10.3 grams of CH4 are burned completely, how much heat will be produced?

10. 3 g CH4

16.05 g CH4

1 mol CH4

1 mol CH4

802.2 kJ

= 514 kJ

ΔH = -514 kJ, which means the heat is released for the reaction of 10.3 grams CH4

Ratio from balanced equation

1

Start with known valueConvert to moles Convert moles to desired unit

section 15-1 the nature of energy energy is the ability to do work or produce heat.energy...

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