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
Exists in 2 forms:
Kinetic energy – energy of motion
Potential energy – energy at rest or energy of position
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

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 = J
1kcal = 4186J
1 J = cal

6. Heat Capacity
The amount of heat it takes to change an object’s temperature by 1ºC
Depends on an object’s mass
Ex. A cup of water has a greater heat capacity than a drop of water.

7. 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)

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

9. 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.

10. Example
C= q/ (m∆T)
C=114 J/ (10.0 g x 25.4°C)
C = 0.45 J/g C

11. 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

12. 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°C)
q = J

13. 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

14. 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 = C
T = Tf-Ti
= Tf-25
Tf = C