Heat and Thermodynamics

Review What are the three phases of matter?

The Phases There are three basic phases of matter: solid, liquid, and gas. We can change between these phases: A. Solid changing to a gas is called sublimation B. Liquid changing to a gas is called evaporation C. Gas changing to a liquid is called condensation D. Solid changing to a liquid is called melting E. Liquid changing to a solid is called freezing F. Gas changing to a solid is called deposition

The Phases At certain temperatures matter can be in multiple forms H 2 O (l)  H 2 O (s) is the chemical equation for water when it is melting or freezing. Pure water has a melting/freezing point of 0°C. Above 0°C melting will happen & the equation goes from right to left. Below 0°C freezing will happen & the equation goes from left to right.

The Experiment You have two beakers on hot plates. One beaker contains one ice cube. The other beaker contains eight ice cubes. For each beaker, the temperature and the time were recorded as the ice cubes melted.

The Data

The Graphical Representation

What does this all mean? Segment 1 (melting point): At the solid surface melting is occurring as molecules slip by one another. Energy is used for the phase change at the solid surface so temperature stays constant. The type of particle motion is changing, not the speed of the particles.

What does this all mean? Segment 2 (liquid heating): All molecules are moving past each other. All the increasing heat energy is used to increase the velocity of the particles. Kinetic energy increases so temperature goes up.

What does this all mean? Segment 3 (boiling point): Energy is used to make more & more of the particles escape the liquid. Escaping particles form water vapor. Energy is used for the phase change at the liquid surface so temperature stays constant.

What does this all mean? The Slope: Why is the slope of the curve larger for the beaker with 1 ice cube than for the beaker with 8 ice cubes? There are more particles in the beaker so it will take longer for them to melt & heat. Temperature: The temperatures at melting & boiling were the same for both beakers. Temperature is a measure of the average kinetic energy of the particles in a substance.

What does this all mean? Heat: The beaker with 8 ice cubes needed more heat to reach the melting & boiling points because it had more particles. Heat is a measure of the total amount of energy transferred from an object of high temperature to low temperature.

Kinetic Energy KE = ½m x v 2 M stands for mass V stands for velocity KE = ½m x v 2 = (kg) x (m/s 2 ) = Joules Joules (kgm/s 2 ) are the unit for energy.

Heat Capacity Specific heat capacity is the amount of energy needed to raise the temperature of 1 gram of a substance by 1°C. The lower the specific heat of a substance, the faster the object heats up. Specific heat capacity is abbreviated C p & has units of J/g°C.

What does a joule of energy look like? 1 gram of pure ethyl alcohol produces 29,700 joules of energy The average 18 year old creates 14,000 joules of energy per day. This energy keeps your body at 98.6º F.

How do I calculate how much heat an object holds? The Greek letter delta, Δ, means change in. ΔT means change in temperature. Change in temperature is usually found by taking the final T & subtracting the initial T ΔT = Tfinal – Tinitial = T2 – T1

Cont.. The formula for calculating change in heat of a certain mass of a substance in a given phase (s, l, g) through a temperature range is: H = mC p ΔT H= heat; m= mass; C p = specific heat; ΔT = change in temperature The units are (Grams) (°C) (J/g°C) = Joules (g) (°C) ( J ) = J g°C

Example 1 Calculate the joules of energy required to heat 454 g of water from 5.4°C to 98.6°C. The specific heat of liquid water is J/gºC. H = mC p ΔT H = 454 g * J * (98.6 ºC – 5.4ºC) = 93.2ºC 1 gºC 1 H = J

Example 2 A 2.80 g sample of pure metal requires 10.1 J of energy to change temperature from 21ºC to 36ºC. What is the metal?

Example 2- solution H = mC p ΔT 10.1 J = 2.80 g * C p * (36ºC – 21ºC) 10.1 J = 2.80 g * C p * 15ºC 0.240= C p C p = J/g°C SILVER SPECIFIC HEAT VALUES SubstanceSpecific Heat Capacity (J/g°C) Aluminum (s) 0.89 Iron (s) 0.45 Mercury (l) 0.14 Carbon (s) 0.71 Silver (s) 0.24 Gold (s) 0.13 Tungsten (s) Copper (s) 0.385

Define the term exothermic. Heat or energy is released- EXiting Define the term endothermic. Heat or energy is absorbed- ENtering

Δ H Heat of fusion (ΔH fus ) is the energy needed to melt 1 mole of a substance or joules needed to melt 1 gram. Heat of vaporization (ΔH vap ) is the energy in kJ needed to vaporize 1 mole of a substance or joules needed to vaporize 1 gram.

Equations for Heat FILL in your diagram evaporation condensation melting freezing - H in J + H in J Cooling Curve Heating Curve

Write this on the back of your diagram Words to watch for:  Is heated  Is cooled  “ from ______  “ to ________  “ through _______  Heat energy is released  Heat energy is absorbed This means:  Up the stairs = +J, Exo  Down the stairs = -J Endo  Start here  End here  INCLUDE this step  Exo, + J  Endo, - J

Example 3 Find the heat released as 10.5g H 2 O cools from 130°C to -15°C. Step 1- Fill in MP and BP for H 2 O 0 °C 100 °C 130 °C -15 °C

Given information: MP = 0°C BP = 100°C H fus = 334 J/g H vap = 2260J/g C p solid = 2.07 J/g°C C p liquid = 4.18 J/g°C C p gas = 2.04 J/g°C

Example 3: Solution Step 1: Find the heat needed to be released from gaseous (steam) H 2 O to reduce the temperature of the gas. m * C p(gas) * ΔT ; ΔT= T final – T inital 10.5 g * 2.04 J * (100 °C – 130 °C) 1 gºC 10.5 g * 2.04 J * -30 °C = J 1 gºC

Example 3: Solution Step 2: Find the amount of heat that needs to be released to condense the steam into a liquid. m * H vap 10.5 g * 2260 J = J 1 g H vap is – because we are doing DOWN the stairs

Example 3: Solution Step 3: Calculate how much heat needs to be released to decrease the temperature of the liquid water. m* C p(liquid) * ΔT 10.5 g * 4.18 J * (0 °C -100 °C ) 1 gºC 10.5 g * 4.18 J * -100 °C = J 1 gºC

Example 3: Solution Step 4: Calculate the heat that needs to be released to freeze all of the water. m * H fus 10.5 g * 334 J = J 1 g H fus is – because we are doing DOWN the stairs

Step 5: Calculate how much heat must be released to decrease the temperature of the solid water from100 ºC to -15ºC. m * C p(solid) * ΔT 10.5g * 2.07 J * (-15ºC - 0ºC) 1 gºC 10.5 g * 2.07 J * -15ºC = J 1 gºC

Example 3: Solution Step 6: Add up all of the products to get a total J J J J J J This heat needs to be released to decrease water from 130ºC to -15ºC.

Cooling Curve vs Heating Curve When changing a substance from an ice to gas, then a heating curve is used instead of a cooling curve. Uses the same process as a cooling curve, but backwards and the answer is positive. - H in J + H in J Heating Curve Cooling Curve

Determine the final temperature of a 15.5 g block of solid copper originally at 30°C if you apply 680 J of energy to it. Hint- Use the equation: H = m C p ∆ T ∆ T = (T final – T initial ) °C

Calorimetry Calorimeter: a very insulated device used to measure heat energy. A reaction happens in the container & the heat given off raises the temperature of the water. (mass) (ΔT) (Cp water liquid ) = heat (joules) Calorimeters are used to measure food energy. Nutritional information is determined in a lab. A calorie is the energy needed to raise 1 gram of H 2 O by 1°C. 1 calorie = 4.18 Joules

Enthalpy Enthalpy is just another word for heat. The symbol for enthalpy is ΔH. Enthalpy is the total energy a substance contains. ΔH rxn is the symbol for heat of reaction. Heat of reaction is the amount of energy (kJ) released or absorbed during a reaction. ΔH f is the symbol for heat of formation. Heat of formation is the amount of energy absorbed or released during the synthesis of 1 mole of a compound.

Endothermic or Exothermic? This graph shows an exothermic reaction. Energy is released or given off to the surroundings. Enthalpy or ΔH is negative during an exothermic reaction because heat is going out of the substance. If you could feel the reaction happening, it would feel warm/hot.

Endothermic or Exothermic? This graph shows an endothermic reaction. Energy is absorbed or taken in from the surroundings. Enthalpy or ΔH is positive during an endothermic reaction because heat is going into the substance. If you could feel the reaction happening, it would feel cold.

Activation Energy Activation energy is the amount of energy needed to start a reaction. Catalysts are used to lower activation energy. How does this help? 1. Don’t need as much energy to get the reaction started. 2. The reaction will take less time, speeding it up!  Example: Enzymes (amylase, lactase, lipase)

Hess’s Law The total enthalpy change for a chemical or physical change is the same no matter how many steps it takes to occur.

Entropy Entropy is the degree of disorder or randomness of a system. Entropy is represented by the symbol ΔS.  More disorder has a positive ΔS.  Less disorder has a negative ΔS.

If we go from solid  liquid  gas entropy is positive because the molecules are more disorganized If we go from gas  liquid  solid entropy is negative because the molecules more organized.

Free Energy Free energy a quantity that combines both enthalpy and entropy. It is the amount of potential or stored energy available to do work or cause a change. The symbol for free energy is ΔG. Example: diffusion  Spray perfume or cologne and the molecules will slowly travel through out the room naturally.

Free Energy By comparing enthalpy and entropy as well as temperature it can be determined how much free energy a substance has.

Free Energy Free energy can also tell us if a reaction is spontaneous.  Spontaneous means that a reaction will occur automatically, (not necessarily quickly). If ΔG is negative the reaction is spontaneous. If ΔG is positive the reaction is nonspontaneous. How do I know what the sign of ΔG is?  Use the equation: ΔG = ΔH – TΔS

Example 4 If ΔH is +12, ΔS is -4.10, & T is 100 K, what is ΔG? Is this reaction spontaneous or nonspontaneous?

Example 4 - Solution ΔG = ΔH – TΔS ΔG = +12 – (100 ° x -4.10) ΔG = +12 – -410 ΔG = 422 Spontaneous or Nonspontaneous? The End!