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Chemistry

CHM135H1

Dwayne Miller

Winter

Description

Chapter 6
Exam Preparation:
Concepts/Skills – Check List
Chapter 6
- First Law of Thermodynamics
The first law of thermodynamics (law of conservation of energy) states that the
total energy in the universe is conserved.
= + = 0
universe system surroundings
- Definition of a State Function
State function: the internal energy of a system, dependent on the current state
(composition, volume, pressure and temperature) and not the path taken to reach
the state.
- Definition of System and Surroundings
System: part of the universe that we are focusing on
Surrounding: everything in the universe not including the system
- Sign convention from the System view
Energy Transferred as Heat Only
1. Heat flowing out of the system: q = negative and = negative
2. Heat flowing into the system: q = positive and = positive
Energy Transferred as Work Only
1. Work done by the system: q = negative and = negative
2. Work done on the system: q = positive and = positive
q w
+
+ + +
+ - Depends on size of q
and w
- + Depends on size of q
and w
- - -
- Use of Hess’ Law of Heat Summation
Hess’ Law: the enthalpy change of an overall process is the sum of the enthalpy
changes in the individual steps
overall 1+ 2+ 3 + …
- Be able to calculate heat released given number of moles, mass, or be
able convert heat to number of moles generated (all permutations)
Sample Problem 6.3
Finding the Quantity of Heat from a Temperature Change
Problem A layer of copper welded to the bottom of a skillet weighs 125 g. How
much heat is needed to raise the temperature of the copper layer from 25°C to
300.°C? The specific heat capacity (c) of Cu is given in Table 6.2.
Plan We know the mass (125 g) and c (0.387 J/g·K) of Cu and can find ΔT in °C,
which equals ΔT in K. We then use Equation 6.7 to calculate the heat.AT T
Tinitial
final
300 °C
25°C
275°C
275 K
q c x mass (g) x AT 0.387 Jlg.K X 125 g x 275 K
1.33x 104 J
water
final
nitial
(28.49°C
25.10PC) 3.39°C
3.39 K
AT
AT
solid final nitial
(28.490C
100.00 C)
E 71.51°C
71.51 K
CHo
x massilo xATilo 4.184 J/g K x 50.00 g x 3.39 K
0.450 J/g K
solid
massed X ATeed
22.05 g X (-71.51 K)
AT T Tinitial final 300 °C 25°C 275°C 275 K q c x mass (g) x AT 0.387 Jlg.K X 125 g x 275 K 1.33x 104 J water final nitial (28.49°C 25.10PC) 3.39°C 3.39 K AT AT solid final nitial (28.490C 100.00 C) E 71.51°C 71.51 K CHo x massilo xATilo 4.184 J/g K x 50.00 g x 3.39 K 0.450 J/g K solid massed X ATeed 22.05 g X (-71.51 K)Solution Calculating ΔT and then q:
Determining the Specific Heat Capacity of a Solid
Problem You heat 22.05 g of a solid in a test tube to 100.00°C and add it to
50.00 g of water in a coffee‐ cup calorimeter. The water temperature changes
from 25.108C to 28.49°C. Find the specific heat capacity of the solid.
Plan We are given the masses of the solid (22.05 g) and of H O (50.00 g2, and
we can find the temperature changes of the water and of the solid by subtracting
the given values, always using T final T initialsing Equation 6.7, we set the heat
released by the solid (−q solidequal to the heat absorbed by the water (q water). The
specific heat of water is known, and we solve for c solid
Solution Finding ΔT solidnd ΔT water
Solving for c solid
Determining the Enthalpy Change of an Aqueous Reaction
Problem You place 50.0 mL of 0.500 M NaOH in a coffee‐ cup calorimeter at
25.00°C and add 25.0 mL of 0.500 M HCl,

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