THERMODYMANICS

1. THERMODYMANICS Thermodynamics is the study of the motion of heat energy as it is transferred from the system to the surrounding or from the surrounding to the system. The transfer of heat could be due to a physical change or a chemical change. There are three laws of chemical thermodynamics.

2. CHEMICAL THERMODYMANICS The first law of thermodynamics: Energy and matter can be neither created nor destroyed; only transformed from one form to another. The energy and matter of the universe is constant. The second law of thermodynamics: In any spontaneous process there is always an increase in the entropy of the universe. The entropy is increasing. The third law of thermodynamics: The entropy of a perfect crystal at 0 K is zero. There is no molecular motion at absolute 0 K.

3. HEAT The energy that flows into or out of a system because of a difference in temperature between the thermodynamic system and its surrounding. Symbolized by "q". • When heat is evolved by a system, energy is lost and "q” is negative (-). • When heat is absorbed by the system, the energy is added and "q" is positive (+).

4. HEAT FLOW Heat can flow in one of two directions: Exothermic To give off heat; energy is lost from the system: (-q) Endothermic To absorb heat; energy is added to the system: (+q)

5. If the heat transfer involves a chemical reaction then q is called:HEAT OF REACTION The heat energy (DH; enthalpy) required to return a system to the given temperature at the completion of the reaction. q = DH at constant pressure The heat of reaction can be specific to a reaction like: HEAT OF COMBUSTION The quantity of heat energy given off when a specified amount of substance burns in oxygen. UNITS: kJ/mol (kilojoules per mole) or kcal/mol (kilocalories per mole)

6. HEAT CAPACITY & SPECIFIC HEAT HEAT CAPACITY: The quantity of heat needed to raise the temperature of a substance one degree Celsius (or one Kelvin). q = CpDT SPECIFIC HEAT: The quantity of heat required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). q = s x m x DT Both Cp & s are chemical specific constants found in the textbook or CRC Handbook.

7. UNITS for HEAT ENERGY Heat energy is usually measured in either Joules, given by the unit (J), and kilojoules (kJ) or in calories, written shorthand as (cal), and kilocalories (kcal). 1 cal = 4.184 J NOTE: This conversion correlates to the specific heat of water which is 1 cal/g oC or 4.184 J/g oC.

8. SPECIFIC HEAT • Determine the energy (in kJ) required to raise the temperature of 100.0 g of water from 20.0 oC to 85.0 oC? m = 100.0 g DT = Tf -Ti = 85.0 - 20.0 oC = 65.0 oC q = m x s x DT s (H2O) = 4.184 J/ g - oC q = (100.0 g) x (4.184 J/g-oC) x (65.0oC) q = 27196 J (1 kJ / 1000J) = 27.2 kJ • Determine the specific heat of an unknown metal that required 2.56 kcal of heat to raise the temperature of 150.00 g from 15.0 oC to 200.0 oC? S = 0.0923 cal /g -oC

9. LAW OF CONSERVATION OF ENERGY • The law of conservation of energy (the first law of thermodynamics), when related to heat transfer between two objects, can be stated as: The heat lost by the hot object = the heat gained by the cold object -qhot = qcold -mh x sh x DTh = mc x sc x DTc where DT = Tfinal - Tinitial

10. LAW OF CONSERVATION OF ENERGY • Assuming no heat is lost, what mass of cold water at 0.00oC is needed to cool 100.0 g of water at 97.6oC to 12.0 oC? -mh x sh x DTh = mc x sc x DTc - (100.0g) (1 cal/goC) (12.0-97.6oC) = m (1 cal/goC) (12.0 - 0.0 oC) 8560 cal = m (12.0 cal/g) m = 8560 cal / (12.0 cal/g) m =713 g • Calculate the specific heat of an unknown metal if a 92.00 g piece at 100.0oC is dropped into 175.0 mL of water at 17.8 oC. The final temperature of the mixture was 39.4oC. s (metal) = 0.678 cal/g oC

11. PRACTICE PROBLEM #7 180. J 1. Iron metal has a specific heat of 0.449 J/goC. How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 oC, when it is placed in a beaker of boiling water at 1 atm? 2. How many calories of energy are given off to lower the temperature of 100.0 g of iron from 150.0 oC to 35.0 oC? 3. If 3.47 kJ were absorbed by 75.0 g H2O at 20.0 oC, what would be the final temperature of the water? 4. A 100. g sample of water at 25.3 oC was placed in a calorimeter. 45.0 g of lead shots (at 100 oC) was added to the calorimeter and the final temperature of the mixture was 34.4 oC. What is the specific heat of lead? 5. A 17.9 g sample of unknown metal was heated to 48.31 oC. It was then added to 28.05 g of water in an insulted cup. The water temperature rose from 21.04 oC to 23.98oC. What is the specific heat of the metal in J/goC? 1.23 x 103 cal 31.1 oC 1.28 J/g oC 0.792J/goC

12. GROUP STUDY PROBLEM #7 • _____1. A 250.0 g metal bar requires 5.866 kJ to change its temperature from 22.0oC to 100.0oC. What is the specific heat of the metal in J/goC? • _____2. How many joules are required to lower the temperature of 100.0 g of iron from 75.0 oC to 25.0 oC? • _____ 3. If 40.0 kJ were absorbed by 500.0 g H2O at 10.0 oC, what would be the final temperature of the water? • _____ 4. A 250 g of water at 376.3 oC is mixed with 350.0 mL of water at 5.0 oC. Calculate the final temperature of the mixture. • _____5. A 400 g piece of gold at 500.0oC is dropped into 15.0 L of water at 22.0oC. The specific heat of gold is 0.131 J/goC or 0.0312 cal/goC. Calculate the final temperature of the mixture assuming no heat is lost to the surroundings.