Download CCD Apologia Chemistry Syllabus 2011-12

Apologia Chemistry – Module 2 Pre-Lab Preparation - KEY
Module #2: Energy, Heat and Temperature
Pre-Lab Prep: Key Definitions and Math Necessary for these Labs
Define Energy: The ability to do work – to cause motion.
Work: The force applied to an object times the distance that the object travels parallel to that
force. For work to occur, there must be motion.
Heat: Energy transferred as a consequence of temperature differences. It is energy on its way
from a hot body to a cool body.
First Law of Thermodynamics: Energy cannot be created or destroyed. It can only change form.
Potential energy: Stored energy
Kinetic energy: Energy in motion. Heat is a form of kinetic energy.
Metric unit for energy is the: Joule
Celsius temperature standard is defined by the freezing and boiling temperature of water which is
0 oC and 100 oC respectively.
The formula to convert Fahrenheit scale into Celsius is:
Since the formula to convert Celsius into Fahrenheit is
then algebraically we can say (9/5)(oC) = oF – 32
oF
= (9/5)(oC) + 32
and finally
oC
= (5/9)( oF – 32)
Normal body temperature is 98.6 oF – what is this temperature in Celsius?
5/9 * (98.6 – 32) = 37 oC
What is the boiling temperature of water in Fahrenheit?
((9/5) * 100) + 32 = 212 oF
Definition of Kelvin:
K = oC + 273.15
oF
equivalent of 0.00o Kelvin is: (5/9)( oF – 32) = -273.15  ( oF – 32) = -273.15 -459.67
Define calorie: The amount of heat necessary to warm one gram of water one degree Celsius
Calorie scale in Joules:
food/chemistry:
1 calorie = 4.185 Joules
1 food calorie (Cal) = 1,000 chemistry calories (cal)
Absorbed or released heat is measured by the equation: q = m*c*ΔT
Apologia Chemistry – Module 2 Pre-Lab Preparation - KEY
Define specific heat capacity (“c” in the previous equation) = The amount of heat necessary to
raise the temperature of 1 gram of a substance by 1 degree Celsius.
BE SURE YOU fully understand the examples on p. 51 – 53 and can work a similar problem. Try
at least one of the OYO problems:
2.6 A 502 g. iron rod is heated to an unknown initial temperature. If, while cooling down to a
temperature of 22 oC the rod releases 597.5 Joules of energy, what was its initial temperature?
(HINT: When an object loses energy, its “q” is negative.)
From q = m*c* ΔT what DON’T we know: ΔT
-597.5 J / g oC = 502 g. *0.4521 oC * ΔT
ΔT = ________-597.5 J______ = -2.633 oC
502 g. *0.4521 J/g oC
Initial temp = 22+2.633 = 24.633 = 25 oC
2.7 If a 15.6 g. object requires 836.8 J of heat to raise its temperature by 21 oC, what is the specific
heat of the object?
From q = m*c* ΔT what DON’T we know: c
836.8 J = 15.6g * c * 21 oC
c = ______836.8 J______
15.6 g. *21 oC
= 2.554 J/g oC
Specific heat of water, in calories and Joules, is (p. 53)
1.000 cal
g oC
or
4.184 __ J__
g oC
BE SURE YOU fully understand the two examples on p. 55 - 58. Work Example 2.5 yourself:
A 28.5-gram block of an unknown metal is heated to a temperature of 151.7 oC. It is then dropped
into a calorimeter that has a specific heat of 1.95 (J/g oC) and is filled with a mass of 150 grams of
water. If the initial temperature of the water was 22.0 oC, and its final temperature was 24.2 oC,
what is the specific heat of the metal and what is its identity? Using q = m*c* ΔT and -qobject =
qwater + qcalorimeter
ΔT = 24.2 – 22.0 = -2.2 oC
qcalorimeter = (50g)*(1.95 J/g oC)*(2.2 oC) = 21 J
qwater = (150.0g)*(4.184 J/g oC)*(2.2 oC) = 1,400 J
-qmetal = qwater + qcalorimeter
= -qmetal
= 21 + 1400
qmetal = -1421 J
ΔTmetal = 24.2 – 151.7 = -127.5 oC
c = _________-1400 J______
28.5 g. *(-127.5 oC)
= 0.39 J/g oC  copper based on p. 50