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. Disorder 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. SPECIFIC HEAT SPECIFIC HEAT: 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 = C p x m x  T Specific heat is measured under constant pressure. Common specific heats are listed on pg 513.

6. 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 o C or 4.184 J/g o C.

7. SPECIFIC HEAT  Determine the energy (in kJ) required to raise the temperature of 100.0 g of water from 20.0 o C to 85.0 o C? m = 100.0 g  T = T f -T i = 85.0 - 20.0 o C = 65.0 o C q = m x s x  Ts (H 2 O) = 4.184 J/ g - o C q = (100.0 g) x (4.184 J/g- o C) x (65.0 o C) 27.2 kJ 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 o C to 200.0 o C? S = 0.0923 cal /g - o C

8. Calorimetry  Study of heat transfers  A calorimeter is an instrument used to seal in a reaction so that no heat is lost.

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 -q hot = q cold -m h x C P h x  T h = m c x C P c x  T c where  T = T final - T initial

10. LAW OF CONSERVATION OF ENERGY  Assuming no heat is lost, what mass of cold water at 0.00 o C is needed to cool 100.0 g of water at 97.6 o C to 12.0 o C? -m h x C P h x  T h = m c x C P c x  T c - (100.0g) (1 cal/g o C) (12.0-97.6 o C) = m (1 cal/g o C) (12.0 - 0.0 o C) 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 100.0 o C is dropped into 175.0 mL of water at 17.8 o C. The final temperature of the mixture was 39.4 o C. (metal) C P (metal)

11. PRACTICE PROBLEMs 1. Iron metal has a specific heat of 0.449 J/g o C. How much heat is transferred to a 5.00 g piece of iron, initially at 20.0 o C, 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 o C to 35.0 o C? 3. If 3.47 kJ were absorbed by 75.0 g H 2 O at 20.0 o C, what would be the final temperature of the water? 4. A 100. g sample of water at 25.3 o C was placed in a calorimeter. 45.0 g of lead shots (at 100 o C) was added to the calorimeter and the final temperature of the mixture was 34.4 o C. What is the specific heat of lead? 5. A 17.9 g sample of unknown metal was heated to 48.31 o C. It was then added to 28.05 g of water in an insulted cup. The water temperature rose from 21.04 o C to 23.98 o C. What is the specific heat of the metal in J/g o C? 0.792 J/g o C 180. J 1.28 J/g o C 1.23 x 10 3 cal 31.1 o C