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?
Answer
2 NaNO3(s) + H2SO4(l) → Na2SO4(s) + 2 HNO3(g)
 The sulfuric acid is the excess reactant.
 16 g of sulfuric acid - 12.6 g of sulfuric acid =
3.4 g of sulfuric acid in excess
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.
INTRO TO THERMOCHEMISTRY
The study of heat
Objectives
 Today I will be able to:
 Differentiate between temperature and heat
 Explain the process of heat transfer
 Calculate the enthalpy of a system
Temperature vs. Heat
Temperature
Heat
 Average kinetic energy of
 Energy is transferred
molecules in a sample
between two objects due
to temperature difference
 Heat transfer is always
high to low
Direction of heat transfer
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
Endothermic Reaction
 Reaction that absorbs heat
 C + H2O + 113 kJ  CO + H2
Exothermic Reaction
 Reaction that releases heat
 C3H8 + 5 O2  3 CO2 + 4 H2O + 2043 kJ
Measuring enthalpy
 Enthalpy (ΔH) is measured in Joules (J) (SI
unit of heat)
 Equation:
ΔH = m x c
Mass
ΔT
Temperature change
Specific Heat
ΔH = q (in txtbk)
x
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
Note the extremely high specific heat of water.
How does this relate to what you learned in bio?
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.
a. What is the specific heat of this type of
glass?
Answer 1:
 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)
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.
Answer 2
 0.14 J/(g.K)
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?
Answer 3
 329 K
Homework/Classwork
 Complete worksheet