Discovery Lab - Energy

Ch. 17 Thermochemistry and Energy A liquid freezing and boiling at the same time, really?!

Thermochemistry is the study of energy changes that occur during chemical and physical changes

The Nature of Energy Energy is the ability to do work or produce heat Heat, q or Q, is energy; flows due to temperature differences (always hot  cold)

Law of Conservation of Energy states that energy cannot be created nor destroyed; only converted into different types

Kinetic vs. Potential Energy Two main types of energy – kinetic and potential Kinetic – energy of motion Potential – energy due to position or energy stored in chemical bonds Chemical potential energy - the energy stored in a substance because of its composition Example: gasoline

Kinetic vs. Potential Energy

Temperature vs. Heat Temperature is a measure of the motion in a sample Temperature is a measurement of heat. Heat is the total energy of molecular motion, dependent upon amount, size, and type of particles. Heat is energy.

Question Time What is energy? Compare and contrast the two types of energy? What is chemical potential energy? What letter represents heat? Which direction is heat transferred? What is the law of conservation of energy? What is the difference between temperature and heat?

Units of Heat calorie - the amount of heat required to raise the temperature of one gram of pure water by one degree Celsius Calorie – nutritional calorie; 1 Calorie = 1000 calories = 1 kilocalories (kcal) Joule – SI unit of heat 1 calorie = 4.184 J

Converting Energy Units Calorie/calorie/kilocalorie calorie/Joule 1 Calorie = 1000 cal = 1 kcal 1 cal= 4.184 Joule Large nutritional calorie small science calorie small science calorie

Example 1 A cereal has 155 nutritional Calories per serving. How many calories, kilocalories and Joules is this? 155,000 cal 155 kcal 649,000 J

Example 2 A person on a diet consumed 1350 Calories in one day. How many calories, kilocalories and Joules is this? 1,350,000 cal 1,350 kcal Can skip if not enough time. 5,650,000 J

Day 2 – Thermochemistry

Universe Universe = system + surroundings System – the specifc part of universe you wish to study; in chemistry this is your chemical reaction/physical process Surroundings – everything else in the universe When heat is transferred it can flow in or out of the system

Endothermic vs. Exothermic An Exothermic process is one that releases (evolved) heat to its surroundings (feels warm) Eproducts < Ereactants Potential

An Endothermic process is one that absorbs heat from the surroundings (feels cold) Eproducts > Ereactants Potential

Q and heat flow Exothermic process, heat is released, q is negative (-) Endothermic process, heat is absorbed, q is positive (+)

Specific Heat = Cp the amount of heat required to raise the temperature of one gram of that substance by one degree Celsius. Units: J/goC

Specific Heat cont . . . Each substance has a different specific heat Water = 4.184 J/goC Gold = 0.129 J/goC Copper = 0.386 J/goC The lower the specific heat, the lower the amount energy is required to raise its temperature. Does it take more or less energy to raise the temperature in copper compared to water? Less

Calculating Heat Released and Absorbed by a substance Q = m(T)Cp Q = Heat (Joules) m is mass (gram) T (°C) is temperature change Tfinal –Tinitial Cp is specific heat at a constant pressure (J/g °C)

Question Time What is meant by the system? What is meant by the surroundings? What is an endothermic process? What is an exothermic process? What is specific heat? What is the formula to calculate heat?

Example 3-Calculating Heat If the temperature of 56.6 g of ethanol increases from 45.0oC to 80.0oC, how much heat has been absorbed by the ethanol? Specific heat of ethanol = 2.44 J/goC Q = 4830 J

Example 4 -Calculating Heat A 4.00 g sample of a substance was heated from 274 K to 314 K and absorbed 32 J of heat, what is the specific heat of the substance? 0.20 J/gK = Cp

CP Example 5-Calculating Heat If 98,000 J of energy are added to 6200 g of water, what will the change in temperature of the water be? Specific heat of water = 4.184 J/goC 3.8 oC = ∆T

H Example 5-Calculating Heat If 98,000 J of energy are added to 6200 g of water at 14.0oC, what will the final temperature of the water be? Specific heat of water = 4.184 J/goC 3.8 oC = ∆T 3.8 oC = Tf – 14.0 oC 17.8 oC = Tf