CHAPTER 10 ENERGY

10.4 Thermodynamics Thermodynamics: The study of energy. The Law of conservation of energy is also called the First Law of Thermodynamics: The energy of the universe is constant

When we study the heat involved in physical and chemical changes there will be two important factors: Amount of heat involved (number) Direction of heat flow (sign) Endothermic change - heat absorbed from the surroundings (q = +) Exothermic change - heat released to the surroundings (q = -)

To determine the change in energy of a system: E = q + w where E is the amount of energy q represents heat w represents work

10.5 Measuring Energy Changes Thermal energy isn't measurable, but heat is. Measuring Heat involves these units. calorie and Joule calorie: the amount of energy required to raise the temperature of one gram of water one degree Celsius. calorie and Calorie (kcal) are different. Calorie = 1000 calories. As on food labels

Joule can be converted to calories 1 calorie = 4.184 joules convert 60.1 cal into joules 251J convert 34.8 cal into joules 146J convert 47.3J into cal 11.3 cal

Three factors affect how much heat an object absorbs or loses 1. Mass of object 2. Temperature change of object Final temperature minus initial temperature If there is no change in temperature, no heat flows

3. Composition of object Specific heat: energy required to raise the temperature of 1 g of material by 1 Celsius Every substance has a specific heat that is unique to itself. The specific heat of water, for example, is 4.184 J/ g x °C (see table on page 297)

Specific Heat Capacity is used to calculate temperature changes when a substance is heated Q = m ▪ Cp ▪ ΔT Q or q = Heat Energy in Joules m = mass in grams ΔT = change in temperature (Celsius) Cp or s  = specific heat  unit:  joule/g x C

Example: How much heat energy is absorbed when 88 Example: How much heat energy is absorbed when 88.0 g of water is heated from 5.00 oC to 37.0 oC?  ( CP-H2O = 4.184 J/ g x oC) Take inventory first: m = 88.0 g CP = 4.184 J / g x C ΔT = 32 oC ( 37oC - 5oC = 32oC) Q = ? Q = (88.0g)( 4.184J/goC)(32oC) Q = 1.18 x 104 J or 11800 J

Try these examples 100.0 g of water-cools from 30.10°C to 25.05 °C. How much heat is released? (100.0g)(-5.05°C)(4.184 J/g°C) = 2113 J

Q = m (TempFinal – TempInitial) Cp 100.0 g of water at 25.00 °C absorbs 100.0 J of heat. What is its final temperature? Q = m ΔT Cp Q = m (TempFinal – TempInitial) Cp 100.0 J = 100.0 g (TempFinal – 25.00 °C)(4.184J/g °C) TempFinal = 25.24 °C

A stone weighing 2. 0 g absorbs 5. 0 J of heat and warms by 3. 0 °C A stone weighing 2.0 g absorbs 5.0 J of heat and warms by 3.0 °C. What is the specific heat of the stone? What is the heat capacity of the stone? Q = m ΔT Cp Q / m ΔT = Cp 5.0J /2.0g (3.0 °C) = Cp

10.6 Enthalpy = heat of a chemical reaction.