Thermodynamics Energy is... The ability to do work. Conserved.

Made of heat and work. A state function: values that depend on the state of the substance, and not on how that state was reached. Independent of the path, or how you get from point A to B. Work is a force acting over a distance. Heat is energy transferred between objects because of temperature difference. The Universe

Is divided into two halves. The system and the surroundings. The system is the part you are concerned with. The surroundings are the rest.

Exothermic processes release energy to the surroundings. Endothermic processes absorb energy from the surroundings. Units of Heat Calorie (cal) The quantity of heat required to change the temperature of one gram of water by one degree Celsius.

Joule (J) SI unit for heat 1 cal = 4.184 J Direction Every energy measurement has three parts. 1. A unit ( Joules or calories). 2. A number how many (magnitude). 3. A sign to tell direction.

negative - exothermic positive- endothermic Surroundings System Energy DE <0

Surroundings System Energy DE >0 First Law of Thermodynamics

The energy of the universe is constant. Law of conservation of energy. q = heat Take the systems point of view to decide signs. q is negative when the system loses heat energy

q is positive when the system absorbs heat energy Heat Capacity The quantity of heat required to change the temperature of a system by one degree. Three types with different systems: Specific Heat Capacity Molar Heat Capacity (we wont cover this one in Physics) Heat Capacity

Heat Capacity Types Specific heat capacity System is one gram of substance

Units are J/goC q = heat change m = mass C = Specific Heat Capacity DT = = = final temperature initial temperature q = (m)(C)(DT) Heat Capacity Types Heat capacity of an object

Specific for a given amount or unit so it is not mass dependent Units are J/oC q = heat C = heat capacity DT = = = final temperature initial temperature q = CDT Heat Capacity Types

So how do you know which one to use when? Look at the known information Select the value with appropriate units to cancel Example The specific heat of graphite is 0.71 J/gC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

Example The specific heat of graphite is 0.71 J/goC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K. q = mCT = (75,000 g)(0.71 J/gT = (75,000 g)(0.71 J/goC)(348-294)K = 2.88 x 106 J Note that T = (75,000 g)(0.71 J/gT oC and T = (75,000 g)(0.71 J/gT K are interchangeable, since they are the same size.

Example When a piece of copper (5.0 g) is heated for 2.0 seconds, and 100 J of heat energy is transferred to the copper, the temperature increases from 20.0 C to 71.9 C. What is the specific heat of the copper? m = 5.0 g; T = (75,000 g)(0.71 J/gT = (71.9 20.0) = 51.9C; q = 100 J; t = 2.0 s q = mCT = (75,000 g)(0.71 J/gT; q = C = 100 mT = (75,000 g)(0.71 J/gT (5.0)(51.9) time is irrelevant.

= 0.385 J/goC Example If 10.0 g of Cu is heated for 2.0 seconds from 20.0 C and 200 J of heat are absorbed, what is the final temperature of the block? (Lets assume time is again irrelevant.) m = 10.0 g; Ti = 20.0 oC; Tf = ?; CCu = 0.385 J/goC; q = 200 J q = mCT = (75,000 g)(0.71 J/gT; T = (75,000 g)(0.71 J/gT = q/mC = (Tf Ti); Tf = q/mC + Ti

Tf = [200/(10*0.385)] + 20 = 71.9oC Calorimetry Measuring temperature changes to calculate heat changes. Use a calorimeter. Two kinds Constant pressure calorimeter (called a coffee cup calorimeter) Constant volume calorimeter (called a bomb

calorimeter) Calorimetry A coffee cup calorimeter measures temperature and calculates q. An insulated cup, full of water (constant pressure, variable volume) Water is the surroundings in which the system changes Calculate the heat change of water. The specific heat of water is 4.184 J /goC

Heat of water qH2O = CH2O mH2O DT qsystem + qsurroundings = 0 qH2O + qrxn = 0 qH2O = qrxn In interactions between a system and its surroundings the total energy remains constant energy is neither created nor destroyed. Law of Conservation of Energy

Coffee Cup Calorimeter A simple calorimeter. Well insulated and therefore isolated. Measure temperature change. qrxn = qcal Determination of Specific Heat Example

Determining Specific Heat from Experimental Data. Use the data presented on the last slide to calculate the specific heat of lead. q Pb = q H O 2 q H O = mcDT = (50.0 g)(4.184 J/g C)(28.8 22.0)C 2

q H O = 1.4x103 J 2 q Pb = 1.4x103 J = mcDT = (150.0 g)(c)(28.8 100.0)C c Pb = 0.13 Jg1C1 Example Equal masses of liquid A, initially at 100 C, and liquid B, initially at 50 C, are combined in an insulated container. The final temperature of the mixture is 80 C. Which has

the larger specific heat capacity, A or B? mA = mB = m; TiA = 100C; TiB = 50C; Tf = 80C qA = qB; mCAT = (75,000 g)(0.71 J/gTA = mCBT = (75,000 g)(0.71 J/gTB; CAT = (75,000 g)(0.71 J/gTA = CBT = (75,000 g)(0.71 J/gTB; CA/CB = T = (75,000 g)(0.71 J/gTB/T = (75,000 g)(0.71 J/gTA CA/CB = (8050)/(10080) = 1.5. Since CA/CB > 1, CA > CB You do one When 86.7 grams of water at a temperature of 73.0 C is mixed with an unknown mass of water at a temperature of 22.3 C the final temperature of the resulting mixture is 61.7 C. What was the mass of the second sample of water? a. 24.9 g

c. 48.2 g b. 302 g d. 419 g States of Matter General Heating Curve Changes of State for Water

Heat of vaporization (liquid <----> gas) H2O(l) H2O(g) DHv = 2,258 J/g Heat of fusion (solid <----> liquid ) H2O(s) H2O(l) DHf = 333 J/g Heating Curve for Water Total heat absorbed by the water as it is warmed can be

determined by finding the heat involved in each step of the process. q1+ q2 + q3 . = qtotal Note: Cice= Csteam = 2.09 J/goC q = mCDT qphase change = DHphase change

Example Calculate q for the process in which 50.0 g of water is converted from liquid at 10.0C to vapor at 125.0C. Example What is the heat of fusion of lead in J/g if 6.30 kilojoules of heat are required to convert 255 grams of solid lead at its melting point into a a. 0.0250 J/g c. 1.61 J/g

liquid? b. 24.7 J/g d. 40.5 J/g Example What quantity of heat is required to heat 1.00 g of lead from 25 C to the melting point (327 C) and melt all of it? (The specific heat capacity of lead is 0.159 J/g K and it requires 24.7 J/g to convert lead from the solid to the liquid state.)

a. 2.47 J c. 39.4 J b. 48.0 J d. 72.7 J