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

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.

Types of Systems

 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.

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.

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

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

Units of Energy 1 calorie = heat required to raise temp. of 1.00 g of H 2 O by 1.0 o C 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 = joules James Joule

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.

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?

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).

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)]

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

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

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

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 o C = 60.0 o C 2. State what variable you are Required to find.  q = ?

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 J  q = 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.

Let’s try another calculation! If 25.0 g of Al cool from 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 = J/g o C   T = t 2 – t 1 = 37.0 o C o C = o C   q = mc  T  q = (25.0 g)(0.897 J/gK)( o C)  q = J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al to the surroundings = EXOTHERMIC

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)