Regents Chemistry Physical Behavior of Matter Different Phases

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Regents Chemistry Physical Behavior of Matter

Regents Chemistry Physical Behavior of Matter

Different Phases of Matter • An element, compound or mixture can exist in the

Different Phases of Matter • An element, compound or mixture can exist in the form of a solid, liquid or a gas • Solid – rigid form, definite volume and shape, strong attractive forces and crystalline structure • Liquid – not held together as well, can move past one another, no definite shape but definite volume • Gas – minimal attractive forces, no definite shape or volume, expand to shape of container

Other Phases • Vapor – is the gaseous phase of a substance that is

Other Phases • Vapor – is the gaseous phase of a substance that is a liquid or a solid at normal conditions: ex: water vapor • Plasma – is a gas or vapor in which some or all of the electrons have been removed from the atoms. ex: In a planet’s core!

Heating and Cooling Curves • Heating Curves: Constant rate of heating of a substance

Heating and Cooling Curves • Heating Curves: Constant rate of heating of a substance over time – endothermic process!

What Can We Learn From a Heating Curve? • AB: heating of a solid,

What Can We Learn From a Heating Curve? • AB: heating of a solid, one phase present, kinetic energy increases • BC: melting of a solid (melting), two phases present, potential energy increases, kinetic energy remains constant • CD: heating of a liquid, one phase present, kinetic energy increases

What Can We Learn From a Heating Curve? • DE: boiling of a liquid

What Can We Learn From a Heating Curve? • DE: boiling of a liquid (Vaporization), two phases present, potential energy increases, kinetic energy remains constant • EF: heating of a gas, one phase present, kinetic energy increases ***We can tell when the kinetic energy remains constant because the temperature is not increasing!***

Cooling Curves • Shows the constant rate of cooling of a gas at high

Cooling Curves • Shows the constant rate of cooling of a gas at high temperature – an exothermic process

Summary of a Cooling Curve • AB: cooling of a gas (vapor), one phase

Summary of a Cooling Curve • AB: cooling of a gas (vapor), one phase present, kinetic energy decreases • BC: condensation of the gas (vapor) to liquid, two phases present, potential energy decreases, kinetic energy remains constant • CD: cooling of a liquid, one phase present, kinetic energy decreases

Summary of a Cooling Curve • DE: solidification (freezing) of a liquid, two phases

Summary of a Cooling Curve • DE: solidification (freezing) of a liquid, two phases present, potential energy decreases, kinetic energy remains the same • EF: cooling of a solid, one phase present, kinetic energy decreases

Substances That Do Not Follow the Curves • Some substances change directly from a

Substances That Do Not Follow the Curves • Some substances change directly from a solid to a gas – Sublimation – Example: CO 2 changes from a solid to a gas a normal atmospheric pressure • Some substances change directly from gas to a solid – Deposition

Practice Problem Which portions of the graph represent times when heat absorbed and potential

Practice Problem Which portions of the graph represent times when heat absorbed and potential energy increases while kinetic energy remains constant? worksheet

Regents Chemistry • Temperature Scales

Regents Chemistry • Temperature Scales

Temperature Scales ®Celsius ® Based °C on boiling point/freezing point of water ®Kelvin ®

Temperature Scales ®Celsius ® Based °C on boiling point/freezing point of water ®Kelvin ® Based K on absolute zero ®Fahrenheit ® Used °F in U. S. and Great Britain

Conversions ® Key Equations Celsius to Kelvin K = °C + 273 Fahrenheit to

Conversions ® Key Equations Celsius to Kelvin K = °C + 273 Fahrenheit to Celsius °C = 5/9 (°F - 32) Kelvin to Celsius °C = K - 273 Celsius to Fahrenheit °F = 9/5(°C) + 32 **Add the conversions on the right to your worksheet

Practice Problems ® Convert 10 °C to °F °F = 9/5(°C) + 32 =

Practice Problems ® Convert 10 °C to °F °F = 9/5(°C) + 32 = 9/5 (10 °C) + 32 = 50°F Convert 25°C to K K = °C + 273

Regents Chemistry • Measurement of Heat Energy

Regents Chemistry • Measurement of Heat Energy

Energy and Energy Changes ® Energy is the capacity to do work. In other

Energy and Energy Changes ® Energy is the capacity to do work. In other words, it allows us to do things! ® Energy surrounds us and is involved in all of life’s daily functions. ® It comes in many forms!

Energy and Energy Changes ® Energy can be used to change the temperature of

Energy and Energy Changes ® Energy can be used to change the temperature of a substance ® As we heat a substance (put in heat), the vibration of molecules in a substance increases. ® Example: When a solid is heated, the molecules vibrate until they break free and the substance melts.

Specific Heat Capacity • The specific heat capacity of a substance is the amount

Specific Heat Capacity • The specific heat capacity of a substance is the amount of heat required to raise 1 gram of the substance by 1 degree Celsius • For water it is 4. 184 J / g • Compared to other substances, water has a very high specific heat. . what does this mean?

If a substance has a high specific heat capacity… • This simply means that

If a substance has a high specific heat capacity… • This simply means that there will be more energy stored in every 1 gram of water. This also means it will take longer to cool down.

Specific Heat Capacities • Check out the specific heat capacities of different substances!

Specific Heat Capacities • Check out the specific heat capacities of different substances!

Measurement of Heat Energy • Question: Your pool containing 100, 000 grams of water

Measurement of Heat Energy • Question: Your pool containing 100, 000 grams of water absorbs how much heat energy when it warms from 20 °C to 30 °C? • It is easy! We use a formula on our reference tables! q = m. C T

This means what? . . q = m. C T • • q =

This means what? . . q = m. C T • • q = amount of heat absorbed or lost m = mass in grams C = specific heat T = difference in temperature

Back to our problem… • Question: Your pool containing 100, 000 g of water

Back to our problem… • Question: Your pool containing 100, 000 g of water absorbs how much heat energy when it warms from 20 °C to 30 °C? • q = m. C T q = (100, 000 g)(4. 184 J / g • C) (10 °C) = q = 4, 184, 000 Joules!

Rearranging the formula. . • You need to be able to solve for any

Rearranging the formula. . • You need to be able to solve for any of the variables in the equation q = m. C T

Making it easy. . • If we are finding the heat change during the

Making it easy. . • If we are finding the heat change during the melting or boiling phases, we can use the Heat of Fusion or the Heat of Vaporization. . • Why? ? Because temperature remains constant during these periods!

Heat of Fusion and Vaporization • Heat of Fusion – amount of heat energy

Heat of Fusion and Vaporization • Heat of Fusion – amount of heat energy required to melt a unit mass of a substance • For water : HOF = 334 J/g • Heat of Vaporization – amount of energy required to convert a unit mass from liquid to vapor phase • For Water: HOV = 2260 J/g

Practice Problem • How many joules are required to melt 255 g of ice

Practice Problem • How many joules are required to melt 255 g of ice at 0°C? • q = m x Heat of Fusion q = 255 g x 334 J/g = 85, 170 J

Measuring Heat Change ® Calorie = the amount of energy(heat) required to raise the

Measuring Heat Change ® Calorie = the amount of energy(heat) required to raise the temperature of one gram of water by one Celsius degree. ® 1 Calorie (cal) = 4. 184 Joules (J) Metric system SI system

Converting Calories to Joules ® Convert 60. 1 cal of energy into joules 1

Converting Calories to Joules ® Convert 60. 1 cal of energy into joules 1 cal = 4. 184 J 60. 1 cal X 4. 184 J = 1 cal 251 J

Converting Joules to Calories ® Convert 50. 3 J to cal 1 cal =

Converting Joules to Calories ® Convert 50. 3 J to cal 1 cal = 4. 184 J 50. 3 J X 1 cal = 12. 0 cal 4. 184 J

Kilojoules and Kilocalories ® The prefix kilo means 1000 ® energy is often expressed

Kilojoules and Kilocalories ® The prefix kilo means 1000 ® energy is often expressed in kilos because the numbers are large ® We can use Dimensional Analysis to convert. 4. 0 J x 1 k. J 1000 J = 0. 0040 k. J

Converting kilojoules to kilocalories 1 cal = 4. 184 J 1000 kcal = 4184

Converting kilojoules to kilocalories 1 cal = 4. 184 J 1000 kcal = 4184 k. J 500. 0 k. J x 1000 kcal = 2092 kcal 4184 k. J