Heat energy is transferred from a higher temperature object to a lower temperature object until thermal equilibrium is established. YOU NEED YOUR BOOK TODAY

Intro Questions 1.How much heat is lost when 25g of water at 45°C is cooled to 10°C? 2.How much heat is absorbed when 10.0g of ice are melted. 3.If 100.g of 45C water melted the ice from #2, then how much heat did the water lose? 4.To what temperature did the water drop when it released the heat from #3? 5.Is this the final temperature of the water? Why or why not?

Intro Questions In an isolated system, all the heat energy lost by one object is gained by another. Heat transfer can change two factors of an object: its temperature (q = mC p ∆ T) or its phase (q = m ∆ H x ). General Equation for heat transfer: Heat lost by hotter object = heat gained by cooler object -q hot = q cold Depending on what happens to the object tells us what to substitute in for q hot or q cold.

Intro Questions g of copper metal at 90.0°C is added to 150.g of cold 20.0°C water. If the equilibrium temperature ends up to be 22.2°C, what is the specific heat of the metal? q hot + q cold = 0 q Cu + q H2O = 0 What happens to the metal ( ∆ T or ∆ phase?)…which equation do we use to solved for q? q Cu = (mC p ∆ T) Cu

Intro Questions What happens to the water ( ∆ T or ∆ phase?)…which equation do we use to solved for q? q H2O = (mC p ∆ T) H2O (mC p ∆ T) Cu + (mC p ∆ T) H2O = 0 C p of copper = J/g°C Specific heat is the amount of heat needed to raise the temperature of 1g of the substance by 1°C.

Intro Questions 1.What is the final temperature when you add 25.0g of -2.50C ice to 100.g of 75.0C water? q hot + q cold = 0 q H2O + q ice = 0 What happens to the hot water ( ∆ T or ∆ phase?)…which equation do we use to solved for q? q hotH2O = (mC p ∆ T) hotH2O

Intro Questions What happens to the ice ( ∆ T or ∆ phase?)…which equation do we use to solved for q? q ice = (mC p ∆ T) ice q ice = (m ∆ Hfus) ice q coldH2O = (mC p ∆ T) coldH2O q hotH2O + q ice = 0 q hotH2O + q ice + q icemelt + q coldH2O = 0 (mC p ∆ T) hotH2O +(mC p ∆ T) ice +(m ∆ H fus ) ice +(mC p ∆ T) coldH2O = 0

Heat of Reaction There are a few ways to find out the amount of heat gained or lost by a chemical reaction. 1.Bond Energies 2.Hf 3.Hess’s Law 4.Calorimetry

Bond Energy Every bond that exists between atoms requires some amount of energy to break it. Energy is given off when bonds are formed. ∆ H rxn = (∑bonds broken) – (∑bonds formed) So, you need to draw all the molecules, determine how many bonds break and form and use the equation above to determine the heat of reaction.

Bond Energy What is the heat of reaction (kJ/mol) for the combustion of methane? CH 4 + 2O 2  CO 2 + 2H 2 O CH 4 + 2O 2  CO 2 + 2H 2 O ∆ H rxn = -694kJ How much heat is given off when 47.5g of methane is burned? 2050kJ of heat is given off when 47.5g of CH 4 is burned.

Heat of Formation Heat of formation is labeled ∆ H f. You can locate the heats of formation for many compounds in the appendix in the back of your book. The heat of formation of every ELEMENT is 0kJ/mol. ∆ H rxn = (∑ ∆ H f Products) – (∑ ∆ H f Reactants) Much easier than bond energies…just look up the Hf of each compound and plug into the equation.

Heat of Formation What is the heat of reaction of benzene (C 6 H 6 ) in kJ/mol. The ∆ H f for benzene is kJ/mol. You can look up the rest of the values in Appendix L. H2O is in the liquid form in the balanced equation. 2C 6 H 6 + 7O 2  12CO 2 + 6H 2 O (l) ∆ H rxn = (∑ ∆ H f Products) – (∑ ∆ H f Reactants) ∆ H rxn = ( ∆ H f (12CO2) + (6H2O)) – ( ∆ H f (2C6H6) + (0)) ∆ H rxn = (12mol)( kJ/mol) + (6mol)( kJ/mol) – (2mol)(+48.95kJ/mol)

Heat of Formation ∆ H rxn = (12mol)( kJ/mol) + (6mol)( kJ/mol) – (2mol)(+48.95kJ/mol) ∆ H rxn = ( kJ) + ( kJ) – (+97.90kJ) ∆ H rxn = kJ 2C 6 H 6 + 7O 2  12CO 2 + 6H 2 O (l) ∆ H rxn = -6535kJ But, what is ∆ H rxn in kJ/mol C 6 H 6 ? -3268kJ/mol C 6 H 6

Hess’s Law Hess’s Law allows to add up reactions of which we know H rxn for to find the H rxn that we do not know. We want to know the H rxn for C + 2H 2  CH 4 However we are given the following: C + O 2  CO 2 ∆ H rxn = kJ H 2 + ½ O 2  H 2 O ∆ H rxn = kJ CH 4 + 2O 2  CO 2 + H 2 O ∆ H rxn = kJ All are gases except for liquid water and solid carbon.

Calorimetry Just like the last calorimetry problems. q hot + q cold = 0 q hot is the reaction and q cold is the water in the calorimeter. We cannot plug an equation in for q hot, but we know that the water will be changing temperature so… q rxn + (mC p ∆ T) H2O = 0 If we use bomb calorimeter, then the bomb absorbs heat also. q rxn + (C ∆ T) bomb + (mC p ∆ T) H2O = 0

Calorimetry What is H rxn (q rxn ) in kJ/mol if 1.00g of sucrose is burned in a bomb calorimeter. The temperature of g of water in the calorimeter rises from 25.00C to 27.32C. The heat capacity of the bomb is 837J/gC.