1. Thermochemistry Systems, Measuring energy changes, heat capacity and calorimetry.

2. Types of Systems  When considering thermal changes, it is useful to define the involved components of the thermal change into one of 3 types of systems: Open, Closed and Isolated.

3. Types of Systems

4.  When attempting to measure the energy gained or lost by a system, an isolated system is ideal because no energy is lost to the surroundings where measuring the change in temperature is immeasurable.  However an isolated system is impossible since it is not possible to create a substance completely isolated from collisions with surrounding particles.

5. Energy & Chemistry All of thermodynamics depends on the law of CONSERVATION OF ENERGY.  The total energy is unchanged in a chemical reaction.  If PE of products is less than reactants, the difference must be released as KE.

6. Energy Change in Chemical Processes  PE of system dropped. KE increased. Therefore, you often feel a T increase. reactants products

7. Energy Change in Chemical Processes  So, we use changes in T (∆T) to monitor changes in E (∆E).

8. Units of Energy 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 o C. 1000 cal = 1 kilocalorie = 1 kcal 1 kcal = 1 Calorie (a food “calorie”) But we use the unit called the JOULE But we use the unit called the JOULE (Heat (q) is measured in) 1 cal = 4.184 joules James Joule 1818-1889

9. Measuring and Calculating Heat  Measuring the amount of energy that is gained or lost by a system that has undergone a physical, chemical or nuclear change is often a goal of chemists.  The science of measuring changes in energy is called Calorimetry. It takes its name from an older unit of energy, the Calorie.

10. Heat Capacity The heat capacity of a substance is the amount of heat energy (q) it must consume in order to raise its temperature by 1K or 1ºC. Which has the larger heat capacity?

11. Specific Heat Capacity  Not all substances are the same in terms of the amount of thermal energy they can store.  Every pure substance involved in a chemical reaction has a unique heat capacity  The amount of energy that is required to raise the temperature of one gram of a given substance is referred to as its specific heat capacity (J/g-K or J/g-ºC).

12. Specific Heat Capacity How much energy is transferred due to a change in temperature (∆T)? a change in temperature (∆T)? The heat “lost” or “gained” is related to The heat (q) “lost” or “gained” is related to 1. sample mass (m) 2. change in T (∆T) 3. Specific Heat Capacity (c) [q = -ve  exo (syst  surr), q = +ve  endo (surr  syst)]

13. Note the tremendous difference in Specific Heat. Water’s value is VERY HIGH.

14. Water is Industrious!  For water, c = 4.18 J/(g o C) in Joules, and c = 1.00 cal/(g o C) in calories.  Thus, for water:  it takes a long time to heat up, and  it takes a long time to cool off!  Water is used as a coolant! 14

15. Molar vs Specific Heat Capacity  The heat capacity of 1 mol of a pure substance is known as its molar heat capacity (J/mol-K or J/mol-ºC).  The heat capacity of 1 gram of a substance is known as its specific heat capacity (J/g-K or J/g-ºC). A useful relationship is J X g = J g o C mole mol o C or specific heat X molar mass = molar heat capacity

16. Let’s try a calculation o C o C  How much energy is required to heat 2.0 L of water from 20.0 o C to 80.0 o C? 1. State what variables you know first (Givens). Since the density of water if 1.0 g/mL, the volume of water is an equivalent amount in grams.  m H2O = 2.0 L = 2.0 kg = 2000 g = 2.0 x 10 3 g o C  c = 4.18 J/g o C   T = t 2 – t 1 = 80.0 o C - 20.0 o C = 60.0 o C 2. State what variable you are Required to find.  q = ?

17. Let’s try a calculation 3. Analyze your variables:  Are all your units correct? Do you need to convert any?  What equation do you need to solve this?   q = mc  T 4. Solve the equation.   q = mc  T o C / 60.0 o C  q = 2.0 x 10 3 g x 4.18 J/g o C / 60.0 o C 501 600 J  q = 501 600 J J  q ≈ 5.0 x 10 5 J 5. Paraphrase your answer: It takes J of energy to increase the temperature of 2.0 L or water by 60.0 o C. It takes 5.0 x 10 5 J of energy to increase the temperature of 2.0 L or water by 60.0 o C.

18. Let’s try another calculation! If 25.0 g of Al cool from 310.0 o C to 37.0 o C, how many joules of heat energy are lost by the Al?  m Al = 25.0 g o C  c Al = 0.897 J/g o C   T = t 2 – t 1 = 37.0 o C - 310.0 o C = -273.0 o C   q = mc  T  q = (25.0 g)(0.897 J/gK)(-273.0 o C)  q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al to the surroundings = EXOTHERMIC

19. Recap  Heat Capacity – Energy required to raise the temp of a given quantity of substance by 1.0 ⁰ C or 1.0 K  Specific Heat Capacity – Energy required to raise the temp of one gram of the sub by 1.0 ⁰ C or 1.0 K (J/g ⁰ C or J/gK)  Molar Heat Capacity – Energy required to raise the temp of one mole of the sub by 1.0 ⁰ C or 1.0 K (J/mol ⁰ C or J/molK)