1. Drill __NaNO3(s) + __H2SO4(l) → __Na2SO4(s) + __HNO3(g)
You have 22 grams of sodium nitrate and 16 g of sulfuric acid.
What steps would you take to determine which reactant is in excess?
How would you determine the grams of reactant in excess?

2. Answer 2 NaNO3(s) + H2SO4(l) → Na2SO4(s) + 2 HNO3(g)
The sulfuric acid is the excess reactant.
16 g of sulfuric acid g of sulfuric acid = 3.4 g of sulfuric acid in excess

3. Drill New Quarter – new seats!
What determines the direction of energy transfer between two objects?
Energy moves from hot to cooler objects and continues until the objects reach the same temp.

4. Intro to Thermochemistry
The study of heat
Intro to Thermochemistry

5. Objectives Today I will be able to:
Differentiate between temperature and heat
Explain the process of heat transfer
Calculate the enthalpy of a system

6. Temperature vs. Heat Temperature Heat
Average kinetic energy of molecules in a sample
Energy is transferred between two objects due to temperature difference
Heat transfer is always high to low

7. Direction of heat transfer

8. Enthalpy (ΔH)
Measure of the heat content in a system
As temperature increases, the enthalpy of the system increases
Positive Δ H indicates an endothermic reaction
Negative Δ H indicates an exothermic reaction

9. Endothermic Reaction Reaction that absorbs heat
C + H2O kJ  CO + H2

10. Exothermic Reaction Reaction that releases heat
C3H8 + 5 O2  3 CO2 + 4 H2O kJ

11. Measuring enthalpy
Enthalpy (ΔH) is measured in Joules (J) (SI unit of heat)
Equation:
ΔH = m x c x ΔT
ΔH = q (in txtbk)
Mass
Temperature change
Specific Heat

12. Specific heat
the amount of heat required to raise the temperature of 1 gram of a substance 1 o C.
Depends on:
nature of the material,
mass of the material and
size of the temp change
Look up in a table of values

13. Note the extremely high specific heat of water.
How does this relate to what you learned in bio?

14. Sample Problem 1
A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40. K, and was found to have absorbed 32 J of energy as heat.
What is the specific heat of this type of glass?
Answer: 297 J

15. Answer 1: ΔH = 32 J ΔH = m x c x ΔT or c = ΔH/m x ΔT
Given: m = 4.0 g
ΔT = 40. K
ΔH = 32 J
ΔH = m x c x ΔT or c = ΔH/m x ΔT
C = 32 J/(4.0 g)(40. K) = 0.20 J/(g x K)

16. Sample Problem 2
Determine the specific heat of a material if a 35 g sample absorbed 96 J as it was heated from 293 K to 313 K.

17. Answer 2
0.14 J/(g.K)

18. Sample Problem 3
If 980 kJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be?

19. Answer 3
329 K

20. Homework/Classwork
Complete worksheet