1. Thermochemistry: The study of heat changes that occur during chemical reactions and physical changes of state

2. Energy Energy – the ability to do work or produce heat Heat (q) = the energy that transfers from one object to another because of a temperature difference between them –Heat always flows from a warmer object to a cooler object

3. Energy Kinetic energy – in a chemical reaction temperature is the determining factor The higher the temperature…the faster the particles move…the higher the average kinetic energy Temperature is a measure of the average kinetic energy Kelvin scale: 0 K = -273 °C Potential energy – in a chemical reaction deals with the types of atoms & what bonds they form

4. Law of Conservation of Energy Law of Conservation of Energy – Energy is neither created nor destroyed

5. Heat (q) Heat or energy can be in joules, calories, kilocalories, or kilojoules The SI unit is the joule 1 Cal = 1000 cal = 1 kcal 1 cal = 4.186 J

6. Specific Heat (C) Specific Heat (C) – the amount of heat required to raise 1 gram of a substance by 1  C Specific heat is an intensive property, and therefore does not depend on size Every substance has its own specific heat Ex. Water = 4.18 J/(g x ºC) Glass = 0.50 J/(g x ºC)

7. Specific Heat Units for C = J / g ● ºC (joules per gram degree Celsius) Equation for Specific Heat: C = q / (m Δ T) This equation can be rearranged to solve for heat (q) q= CmΔT C = specific heat; q = heat; m = mass and ΔT = change in temperature

8. Specific Heat A 10.0 g sample of iron changes temperature from 25.0  C to 50.4  C while releasing 114 joules of heat. Calculate the specific heat of iron.

9. Example C= q/ (m∆T) C=114 J/ (10.0 g x 25.4°C) c = 0.449 J/g  C

10. Another example If the temperature of 34.4 g of ethanol increases from 25.0  C to 78.8  C how much heat will be absorbed if the specific heat of the ethanol is 2.44 J/g  C

11. Another example First, rearrange the specific heat formula to solve for heat q = Cm  T q = (2.44 J/g°C)(34.4g)(78.8°C – 25.0°C) q = 4520 J

12. Yet another example 4.50 g of a gold nugget absorbs 276 J of heat. What is the final temperature of the gold if the initial temperature was 25.0  C & the specific heat of the gold is 0.129J/g  C

13. Yet another example C= q/ (m∆T); rearrange to find ∆T = q / (C x m) ∆T = 276 J / (.129 J/g°C x 4.50 g)  T = 475  C  T = Tf-Ti 475 = Tf - 25 Tf = 500  C