THERMODYNAMICS Intro & Calorimetry

Terms Thermodynamics: Energy: Heat: the study of heat changes that occur during chemical changes Energy: the capacity to do work only detected by the effects it causes ex. potential energy, kinetic energy, heat Heat: type of energy & transferred from one object to another can only measure its effects, for example change in temp

System: Surroundings: the part which we focus our attention ex. chemicals Surroundings: everything else ex. water bath, flask, atmosphere Heat can flow from the system to the surroundings OR from the surroundings to the system.

Three types of systems: 1. Open System: Lets energy and matter in an out Ex. Open flask 2. Closed System: Allows energy in and out but not matter Ex. Flask with stopper 3. Isolated System: Allows neither energy nor matter in or out Ex. Calorimeter

Laws of Thermodynamics: No energy is created or destroyed (Law of Conservation of Energy) Heat will flow from a warmer body to a cooler body, until both bodies have the same temp

Exothermic Process: Endothermic Process: a process that releases heat to the surroundings heat “exits” the system “Q” is negative Ex. burning candles, condensation Endothermic Process: a process that absorbs heat from the surroundings heat “enters” the system “Q” is positive Ex. liquid water to water vapor, cooking

Calorimetry Calorimeter: Bomb Calorimeter: a device used to measure the absorption or release of heat in processes Bomb Calorimeter: a device used to measure the heat released by burning a compound

We know that the more heat an object absorbs, the higher its temp can be. The mass of an object and its ability to absorb heat (heat capacity) also affects heat absorption. So, Tf - Ti in oC Q = m c ΔT Heat (J) Specific Heat Capacity (J / goC) Mass (g)

Example: If 125g of copper forms a layer on the bottom of a frying pan, how much heat is needed to raise the temp of the copper from 25oC to 300oC. The specific heat capacity for Cu is 0.39J/goC. Q = m c ΔT m = 125g Ti = 25oC Tf = 300oC c = 0.39J/goC QCu = (125g)(0.39J/goC)( 300oC - 25oC) QCu = 13406.25J or 1.34 x 104J Heat is absorbed by the frying pan.

Recall H2O density of 1g/mL Example: HCl was added to of NaOH in a calorimeter, producing 50mL of water. At the start, the solutions were at 25oC. During the reaction, the highest temp observed was 32oC. What is the heat of the reaction? m = 50g Ti = 25oC Tf = 32oC c = 4.19 J/goC Recall H2O density of 1g/mL So, 50mL x 1g/mL = 50g

Q = m c ΔT QH2O = (50g)(4.19J/goC)( 32oC - 25oC) QH2O = 1466.5 J Heat is absorbed by the water.

Heat is given off by the iron nail Example: Calculate the heat involved when a 5.5g iron nail is cooled from 37oC to 25oC. The heat capacity of iron is 0.45J/goC. Q = m c ΔT QFe = (5.5g)(0.45J/goC)( 25oC - 37oC) QFe = -29.7 J m = 5.5g Ti = 37oC Tf = 25oC c = 0.45 J/goC Heat is given off by the iron nail