Energy and States of Matter (Thermochemistry) Measuring Heat Energy

SURROUNDINGS HEAT HEAT HEAT HEAT SYSTEM SYSTEM EXOTHERMIC ENDOTHERMIC

Heat Energy that flows from something warm to something cooler. (Heat flows from high to low!) A hotter substance gives KE (kinetic energy) to a cooler one When heat is transferred (lost or gained), there is a change in the energy within that substance

HEAT VS. TEMPERATURE Heat & temperature are NOT synonyms Heat is energy that transfers from one object to another because of a temperature difference between the objects. Remember that temperature is defined as a measure of the average kinetic energy of particles in matter So, the Kinetic Energy defines the temp.: Particles moving/vibrating fast = HOT Particles moving/vibrating slow = COLD

HEAT UNITS Joules (J) OR calories (cal) TYPES DEFINITION THE FLOW OF THERMAL ENERGY FROM SOMETHING WITH A HIGHER TEMP TO SOMETHING WITH A LOWER TEMP UNITS Joules (J) OR calories (cal) TYPES THROUGH WATER OR AIR = CONVECTION THROUGH SOLIDS = CONDUCTION TRANSFERRED ENERGY BY COLLISION WITH PHOTON = RADIANT ENERGY

Practice1 A. When you touch ice, heat is transferred from 1) your hand to the ice 2) the ice to your hand B. When you drink a hot cup of coffee, heat is transferred from 1) your mouth to the coffee 2) the coffee to your mouth

Practice2 When you heat 200 g of water for 1 minute, the water temperature rises from 10°C to 18°C. If you heat 400 g of water at 10°C in the same pan with the same amount of heat for 1 minute, what would you expect the final temperature to be? 1) 10 °C 2) 14°C 3) 18°C 400 g 200 g

Practice2 When you heat 200 g of water for 1 minute, the water temperature rises from 10°C to 18°C. If you heat 400 g of water at 10°C in the same pan with the same amount of heat for 1 minute, what would you expect the final temperature to be? 1) 10 °C 2) 14°C 3) 18°C 400 g 200 g

Reasoning Heating twice the mass of water using the same amount of heat will raise the temperature only half as much. 400 g 200 g

Some Equalities for Heat Heat is measured in calories or joules 1 kcal = 1000 cal 1 calorie = 4.18 J 1 kJ = 1000 J

calories and Calories A calorie is defined as the quantity of heat needed to raise the temperature of 1 gram of pure water by 1oC Abbreviation = cal with small c. There is a difference between a calorie and a Calorie calorie is defined above and is used except when referring to the energy contained in food. Calorie = refers to energy in food 1 Calorie = 1 kilocalorie = 1000 calories ↑ (in food)

Calories and Calories A calorie is defined as the quantity of heat needed to raise the temperature of 1 gram of pure water with 1oC Abbreviation = cal with small c. There is a difference between a calorie and a Calorie calorie is defined above and is used except when referring to the energy contained in food. Calorie = refers to energy in food 1 Calorie = 1 kilocalorie = 1000 calories ↑ (in food)

1 Calorie (nutritional) = 1 kcal 1 Cal = 1000 cal Energy and Nutrition 1 Calorie (nutritional) = 1 kcal 1 Cal = 1000 cal

Joule Joule = is the SI unit of heat and energy named after the English physicist James Prescott Joule (1818-1889) Equivalence statements: 1J = 0.2390 cal 1 cal. = 4.18 J

Specific Heat : c or Cp Why do some foods stay hot longer than others? Why is the beach sand hot, but the water is cool on the same hot day? Why do your scrambled eggs get cold so fast while the pancakes stay warm for much longer?

Specific Heat (Cp) Values Specific heat is the amount of heat needed to raise the temperature of 1 g of a substance by 1°C cal/g°C J/g°C water 1.00 4.18 memorize! aluminum 0.22 0.90 copper 0.093 0.39 silver 0.057 0.24 gold 0.031 0.13

Specific heats are listed on data tables Specific Heat : Cp or c Specific heats are listed on data tables The smaller the specific heat  the less energy it takes the substance to feel hot The larger the specific heat  the more energy it takes to heat a substance up The smaller the specific heat  the less time it takes the substance to cool off The larger the specific heat  the longer time it takes the substance to cool off

Practice3 A. A substance with a large specific heat 1) heats up quickly 2) heats up slowly B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day, and cool at night. Sand must have a 1) high specific heat 2) low specific heat

EXOTHERMIC ENDOTHERMIC

All chemical reactions involve an energy change. The transfer of energy, usually heat, into or out of the reaction mixture. For example: When petroleum burns, heat is given out. When ammonium nitrate dissolves in water, heat is taken in.

Instead of writing ‘enthalpy change’ all the time, chemists use the symbols: Pronounced ‘delta H’. Δ = Greek letter ‘delta’ meaning change. H = enthalpy (see below) So, ΔH means ‘change in enthalpy’. (Enthalpy is the heat content of a system; assuming pressure is constant)

Temperature Changes 1 Some reactions cause the temperature of the reaction mixture to increase. This type of reaction is called exothermic. Heat energy is given out by the reaction so the reaction feels hot.

Temperature Changes 2 Some reactions cause the temperature of the reaction mixture to decrease. This type of reaction is called endothermic. Heat energy is taken in by the reaction so the reaction feels cold.

Representing Energy Changes: The energy changes in a chemical reaction can be conveniently represented using energy level diagrams Energy level diagrams make it easier to decide whether a reaction is exothermic or endothermic See next slide for some examples.

Energy Level Diagrams (copy this!) This energy level diagram shows a reaction where ΔH is positive (an endothermic reaction). This energy level diagram shows a reaction where ΔH is negative (an exothermic reaction). ΔH positive, endothermic. ΔH negative, exothermic.

Making and Breaking Bonds: All chemical reactions involve bond ‘rearrangements’. Bonds are broken and new ones are formed. Chemical reactions are in two stages: 1. Breaking bonds, an endothermic process. 2. Making new bonds, an exothermic process. Breaking chemical bonds is always endothermic. Making chemical bonds is always exothermic.

Summary Exothermic rxns get hot, they give out heat. Chemical reactions happen in two stages: 1. Breaking bonds, an endothermic process. 2. Making new bonds, an exothermic process. Exothermic rxns get hot, they give out heat. Endothermic rxns get cold, they take in heat. (think about that for a second) The energy changes in a chemical reaction can be represented using energy level diagrams and reaction profile diagrams.

Law of Conservation of Energy Law of Conservation of Energy, also called The First Law of Thermodynamics In any chemical or physical process, energy is neither created nor destroyed it is conserved. All the energy involved in a process can be accounted for work, energy, or heat. q aborbed = -q lost

Specific Heat Capacity c or Cp Heat capacity is the amount of heat needed to increase the temperature of an entire object by exactly 1°C q = mc∆T heat = (mass)(specific heat)(change in temperature) units: (grams)( cal OR Joules) (°C) g°C g°C

Heat Capacity Heat capacity is the amount of heat needed to increase the temperature of an entire object by exactly 1°C q = mc∆T heat = (mass)(specific heat)(change in temperature) units: (grams)( cal OR Joules) (°C) g°C g°C Therefore units of heat would be cal or J depending on the specific heat value used.

Heat Calculations q= m x c (of H2O) x T 750 g x 1.00 cal x -28°C g°C A hot-water bottle contains 750 g of water at 65°C. If the water cools to body temperature (37°C), how many calories of heat could be transferred to sore muscles? q= m x c (of H2O) x T 750 g x 1.00 cal x -28°C g°C = -21,000 cal = 21,000 cal

Heat Calculations q= m x c (of H2O) x T 750 g x 4.18 J x -28°C g°C A hot-water bottle contains 750 g of water at 65°C. If the water cools to body temperature (37°C), how many joules of heat could be transferred to sore muscles? q= m x c (of H2O) x T 750 g x 4.18 J x -28°C g°C = -87,760J = 87,760J

Practice4 How many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 1) 60 kcal 2) 7.2 kcal 3) 9.0 kcal 4) 7200 kcal

Solution How many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C?  

Answer How many kcal are needed to raise the temperature of 120 g of water from 15°C to 75°C? 1) 60 kcal 2) 7.2 kcal 3) 9.0 kcal 4) 7200 kcal  

Hsur is the surroundings which is generally water. CALORIMETRY Calorimetry is the process of measuring heat energy Measured using a device called a calorimeter Uses the heat absorbed by H2O to calculate the heat given off by a reaction or an object The amount of heat absorbed (or lost) by the water is equal to the amount of heat released (or gained) by the reaction or object HSYS= -HSUR Hsys is the system or what is taking place in the main chamber (rxn etc.) & Hsur is the surroundings which is generally water.

A COFFEE CUP CALORIMETER USED FOR A REACTION IN WATER, OR JUST A TRANSFER OF HEAT. A BOMB CALORIMETER USED WHEN TRYING TO FIND THE AMOUNT OF HEAT PRODUCED BY BURNING SOMETHING

Phase Changes Quiz

Phase Changes Quiz Draw a diagram showing all 6 of the phase changes for water. Label the 3 states of matter, the names of the phase change processes, and whether each process is exothermic or endothermic.

Phase Change Diagram with names of changes and Exothermic vs. endothermic

*Water’s state of matter changes at certain temps because heat is being gained or lost

*Particles move closer or spread apart more from kinetic energy

Latent Heat of … ΔH Fusion(melting) ΔH Solidification(freezing) ΔH Vaporization ΔH Condensation ΔH Sublimation ΔH Deposition

Latent Heat of … for Water ΔH Fusion(melting) = 80 cal /g or 334 J / g ΔH Solidification = 80 cal /g or 334 J / g ΔH Vaporization = 540 cal /g or 2260 J / g ΔH Condensation = 540 cal /g or 2260 J / g ΔH Sublimation = ? cal /g or ? J / g ΔH Deposition = ? cal /g or ? J / g

Q=mΔHfusion Q=28g (334 J/g) Q=9352 J Calculate the amount of heat in Joules required to melt 28 grams of ice Q=mΔHfusion Q=28g (334 J/g) Q=9352 J

Q=mΔHfreezing (opposite of fusion) Q=155g (334 J/g) Q=51770 J Calculate the amount of heat in Joules needed to freeze 155 grams of water Q=mΔHfreezing (opposite of fusion) Q=155g (334 J/g) Q=51770 J

Calculate the amount of heat in calories needed to vaporize 506 grams of liquid water Q= mΔHvaporization Q= 506g (540 cal/g) Q= 273240 cal

Q= mΔHcondensation (opposite of vaporization) Q=m (540cal/g) Calculate the amount of heat released in calories when 34 grams of water goes through condensation. Q= mΔHcondensation (opposite of vaporization) Q=m (540cal/g) Q = 34g (540 cal/g) = 18360 J

q=6270 J (energy required to raise water to boiling pt.) How much heat in Joules is required to vaporize 50.0g of water that is in a beaker at 70.0°C? q=mcΔT q=(50.0g)(4.18 J/g°C)(100.0°C -70.0°C) q=(50.0g)(4.18 J/g°C)(30.0°C) q=6270 J (energy required to raise water to boiling pt.) Q= mΔHvaporization Q=(50.0g)(2260J/g) Q=113000 J Total Q = 6270J +113000J Total Q = 119270 J

q=(100.0g)(1.00 cal/g°C)(70.0°C – 0.0°C) How many calories are released when 100.0 g of liquid water at 70.0°C freezes completely? q=mcΔT q=(100.0g)(1.00 cal/g°C)(70.0°C – 0.0°C) q=(100.0g)(1.00 cal/g°C)( 70.0°C) q=7000 cal (energy required to lower H2Oto freezing pt.) Q= mΔHsolidification Q=(100.0g)(80 cal/g) Q=8000 cal Total Q = 7000cal +8000cal Total Q = 15000cal