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

C  + 32

F to C

(F - 32)

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 = ?

= 12.7 cm

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)

then: 5 in = 12.7 cm

Example 2:

Convert 5 hours to seconds.

Solution:

5 hrs. = 18,000 sec.

Solution by ratio proportion:

1 hr. = 60 min.

5 hr. = ?

1 min. = 60 sec.= 18,000 sec.

300 min. = ?

Example 3:

Convert 5 mg to kg.

Solution:

1 g = 1000 mg= .005 g

? = 5 mg

1 kg = 1000 g= .000005 kg or 5  10-6 kg

? = .005 g

The second approach is a little easier to follow:

= 5  10-6 kg

Example 4:

Convert 20C to F

Solution:

20C  + 32 = 68F

Example 5:

Convert 14F to C

Solution:

C = (F - 32)

= (14 - 32)

= (-18)

= -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 = (metric)

Equation 2.Specific Heat = (SI)

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) (20 - 5) C

= 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. The specific heat of Fe is

Solution:

Heat = (mass) (sp. ht.) (T)

= (100 g) (40C)

= 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. Given the sp. ht. of water is and sp. ht. of Fe .

Solution:

Heat loss = Heat gained = (mass) (sp. ht.) (T)

Heat Loss = Heat Gained

(40) (.11) (100 - x) = (80) (x - 10) (1)

440 - 4.4x = 80x - 800

-84.4x = -1240

X =

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?
  1. How many calories are required to raise the temperature of 100 g of Fe (sp. ht. from 20C to 90C?
  1. 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?
  1. Calculate the resulting temperature when 200 g of Fe, (sp. ht. at 80C are added to 200 g of water at 20C (sp. ht. Of water .
  1. Calculate the increase in temperature when 1825 J are supplied to 85 g of sand (sp. ht. .787 J g-1C-1)
  1. 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 5D =

Example 10

Calculate the density in g/cm3 and in the SI system if 50 g of a substance occupies a volume of 18 cm3 (ml)

Solution:

D =

= = 2.78 g/ml (metric)

D =

= = = 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 =

= = 10.4 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 =

=

= 1000 ml or 1 L

Equation 6:

Sp. Gr. =

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 IIANSWERS TO SECTION VI

1.4500 mm 1. 229.6 g

2..00005 kg or 5  10-5 kg 2. 840 g

3.96 m 3. 1.5 or 1500 ml

4..05 liters 4. 1.2

5.454.6 g 5. 163.9 g

6.160934.4 cm or 1609.3 m 6. 1.4 g/cm3 1.4  103 kg/m3

7.86F

8.-40F

9.100C

10.-35C

ANSWERS TO SECTION IV

1.2500 cal.

2.770 cal.

3.2 g

4.26C

5.27.3C or 27.3K

6.a. 2750 cal., 11506 J b. 113C