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)