Section I - The Metric, The English, And The Si Systems
Section III- Measurement of Heat
Section V - Density
It will be assumed that every student is familiar with the following units of measurement.
1 Kilometer (km) = 1000 meters (m)
1 Meter (m) = 10 decimeters (dm)
1 Meter (m) = 100 centimeters (cm)
1 Mile = 1.62 km
1 Inch (in) = 2.54 cm
1 Kilogram (kg) = 2.2 lbs.
1 Liter = 1000 milliliters (ml)
1 Liter is approximately 1 quart
For all practical purposes 1 ml = 1 cc
C to F
9
+ 32
5
F to C
5
(F - 32)
9
C 
In converting from one system to another, two approaches are possible. The first approach is ratio and
proportion.
Example 1
Convert 5 inches to centimeters (cm).
Solution:
We know that:
1 in = 2.54 cm
5 in = ?
5 in  2.54 cm
= 12.7 cm
1 in
2nd Approach
When we say that one inch is equal to 2.54 cm then this can be expressed in the following terms:
(1)
or
2.54 cm
in
1 in
2.54 cm
(2)
then:
2.54 cm
 5 in = 12.7 cm
1 in
Example 2:
Convert 5 hours to seconds.
Solution:
5 hrs. 
60 min. 60 sec.
= 18,000 sec.

hr.
min.
Solution by ratio proportion:
1 hr. = 60 min.
5 hr.  60 min.
 300 min.
hr.
5 hr. = ?
1 min. = 60 sec.
300 min.  60 sec.
= 18,000 sec.
1 min.
300 min. = ?
Example 3:
Convert 5 mg to kg.
Solution:
1 g = 1000 mg
?
=
5 mg  1 g
= .005 g
1000 g
5 mg
1 kg = 1000 g
.005 g  1 kg
= .000005 kg or 5  10-6 kg
1000 g
? = .005 g
The second approach is a little easier to follow:
5mg 
1g
1 kg

= 5  10-6 kg
1000 mg 1000 g
Example 4:
Convert 20C to F
Solution:
20C 
9
+ 32 = 68F
5
Example 5:
Convert 14F to C
Solution:
C = (F - 32)
5
9
= (14 - 32)
9
5
= (-18)
9
5
= -10C
SECTION II
Conversion Problems
Convert: 1.
4.5 m to mm
2.
50 mg to kg
3.
105 yds. to meters
4.
50 ml to liters
5.
1 lb. to grams
6.
1 mile to cm.
7.
30C to F
8.
-40C to F
9.
212F to C
10.
-31F toC
SECTION III
MEASUREMENT OF HEAT
Equation 1.
Specific Heat =
calories
(metric)
( C)(g)
Equation 2.
Specific Heat =
joules
(SI)
( K)(g)
Equation 3.
Heat Energy = (mass) (specific heat) (T)
Equation 4.
1 calorie = 4.184 joules
Example 6:
Calculate the number of calories needed to raise the temperature of 10 g of water from 5C to 20C.
Solution:
heat = (10g)
(1 calorie)
(20 - 5) C
( C)(g)
= 150 calories
Example 7:
Calculate the number of calories needed to raise the temperature of 100 g of Fe from 10C to 50C.
.11 cal
.
The specific heat of Fe is
( C)(g)
Solution:
Heat = (mass) (sp. ht.) (T)
(.11 cal)
= (100 g)
(40C)
( C)(g)
= 440 calories
Example 8:
40 g of Fe at 100C were added to 80 g of water at 10C. Calculate the temperature of the mixture.
1 cal.
(.11 cal)
Given the sp. ht. of water is
and sp. ht. of Fe
.
 C. g
( C)(g)
Solution:
100C
Heat loss = Heat gained = (mass) (sp. ht.) (T)
100 - X
X
Heat Loss = Heat Gained
(40) (.11) (100 - x) = (80) (x - 10) (1)
X - 10
10C
440 - 4.4x = 80x - 800
-84.4x = -1240
X=
1240
84.4
X = 14.6C
Example 9:
Calculate the number of joules needed to raise the temperature of 100 g of water from 20C to
51C. (Given sp. ht. of water 4.184 J g-1 C-1)
Solution:
Heat = (m) (sp. ht.) (T)
= (100 g) (4.184 J g-1 C-1) (31C)
= 13,000 J or 1.3 x 104 J
SECTION IV
Energy Problems
1.
How many calories are required to raise the temperature of 50 g of water from 30C to 80C?
2.
How many calories are required to raise the temperature of 100 g of Fe (sp. ht.
.11 cal 
 from 20C
(g)(C) 
to 90C?
3.
The heat of combustion of coal is 6000 cal/g. How many grams are needed to raise the temperature
of 500 g of water from 20C to 44C?
4.
Calculate the resulting temperature when 200 g of Fe, (sp. ht.
of water at 20C (sp. ht. Of water
.11 cal / g 
 at 80C are added to 200 g
C 
1 cal 
.
g  C 
5.
Calculate the increase in temperature when 1825 J are supplied to 85 g of sand (sp. ht. .787 J g -1 C1
)
6.
Calculate the amount of heat needed (in calories and in joules) to raise the temperature of 50 g of
water from 20C to 75C. If this amount of heat is transferred to 1000 g of gold at 25C, what will
the final temperature of the gold be?
SECTION V
DENSITY
Units of density in the metric system is g/cm3 (g/ml) and in the SI system is kg/m3.
Equation 5
D=
M
V
Example 10
Calculate the density in g/cm3 and in the SI system if 50 g of a substance occupies a volume of 18
3
cm (ml)
Solution:
D=
=
D=
M
V
50g
= 2.78 g/ml (metric)
18ml
M (kg)
V (m 3 )
50g
1000g / kg
=
18ml
ml
10 6 3
m
=
50  10 6 kg
18  1000m 3
= 2.78  103kg/m3
Example 9:
Calculate the volume of a solid if its mass is 12.5 g and its density is 1.2 g/ml.
Solution:
From eq. 5
V=
=
M
D
12,5g
= 10.4 ml
12
. g / ml
Example 10:
How many liters will 880 g of a liquid occupy if its density is .88 g/ml?
Solution:
Using V as the subject of Eq. 5
V=
=
M
D
880g
.88g / ml
= 1000 ml or 1 L
Equation 6:
Sp. Gr. =
density of a substance
density of water
SECTION VI
Density Problems
1.
Calculate the mass of a solid if its volume is 20.5 ml and its density is 11.2 g/ml.
2.
The density of a liquid is 1.2 g/ml, calculate the mass of .7 liter.
3.
The density of a solution is 1.22 g/ml, calculate the volume of the solution if its mass is 1830 g.
4.
700 ml of a liquid weighs 840 g. Calculate the sp. gr. of the liquid.
5.
The sp. gr. of chloroform is 1.49. Calculate the mass of 110 ml of the liquid.
6.
A substance that weighs 88 g has a volume of 64 cm3. Calculate the density in g/cm3 and in kg/m3.
ANSWERS TO SECTION II
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
4500 mm
.00005 kg or 5  10-5 kg
96 m
.05 liters
454.6 g
160934.4 cm or 1609.3 m
86F
-40F
100C
-35C
1.
2.
3.
4.
5.
6.
ANSWERS TO SECTION IV
1.
2.
3.
4.
5.
6.
2500 cal.
770 cal.
2g
26C
27.3C or 27.3K
a. 2750 cal., 11506 J
ANSWERS TO SECTION VI
b. 113C
229.6 g
840 g
1.5 or 1500 ml
1.2
163.9 g
1.4 g/cm3
1.4  103 kg/m3