1 problem solving using conversion factors copyright © 2005 by pearson education, inc. publishing...

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Problem Solving using Conversion Factors

Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings

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To solve a conversion problem you need to:• Identify the given unit• Identify the needed unit.

Example: A person has a height of 2.0 meters. What is that height in inches?

The given unit is the initial unit of height. given unit = meters (m)

The needed unit is the unit for the answer. needed unit = inches (in.)

Given and Needed Units

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Learning Check

An injured person loses 0.30 pints of blood. Howmany milliliters of blood would that be?

1. Identify the given and needed units given in thisproblem.

Given unit = _______

Needed unit = _______

pints

milliliters

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2. Write the given and needed units.3. Write a unit plan to convert the given unit to the needed

unit.4. Write equalities and conversion factors that connect the

units.5. Use conversion factors to cancel the given unit and

provide the needed unit.

Unit 1 x Unit 2 = Unit 2Unit 1

Given x Conversion = Needed unit factor unit

contd

# pints = # milliliters

pints milliliters

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Conversion

An injured person loses 0.30 pints of blood. How many milliliters of blood would that be?

0.30 pints x 463 ml = 139 ml 1 pint

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How many minutes are 2.5 hours?

Given unit = 2.5 hr

Needed unit = min

Unit Plan = hr min

Setup problem to cancel hours (hr).

Given Conversion Neededunit factor unit

2.5 hr x 60 min = 150 min (2 SF)

1 hr

Setting up a Problem


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A rattlesnake is 2.44 m long. How many centimeters long is the snake?

1) 2440 cm

2) 244 cm

3) 24.4 cm

Example Problem

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A rattlesnake is 2.44 m long. How many centimeters long is the snake?

Given Conversion Neededunit factor unit

2.44 m x 100 cm = 244 cm

1 m

Solution

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• Often, two or more conversion factors are required to obtain the unit needed for the answer.

Unit 1 Unit 2 Unit 3

• Additional conversion factors are placed in the setup to cancel each preceding unit

Given unit x factor 1 x factor 2 = needed unitUnit 1 x Unit 2 x Unit 3 = Unit 3

Unit 1 Unit 2

Using Two or More Factors

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How many minutes are in 1.4 days?

Given unit: 1.4 days

Factor 1 Factor 2

Plan: days hr min

Set up problem:

1.4 days x 24 hr x 60 min = 2016 min

1 day 1 hr

2 SF Exact Exact = 2 SF

Example: Problem Solving

= 2.0 x 103 min

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• Be sure to check your unit cancellation in the setup.• The units in the conversion factors must cancel to give

the correct unit for the answer.

What is wrong with the following setup?

1.4 day x 1 day x 1 hr 24 hr 60 min

Units = day2/min is not the unit neededUnits don’t cancel properly.

Check the Unit Cancellation

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What is 165 lb in kg?STEP 1 Given 165 lb Need kgSTEP 2 PlanSTEP 3 Equalities/Factors 1 kg = 2.20 lb 2.20 lb and 1 kg

1 kg 2.20 lbSTEP 4 Set Up Problem 165 lb x 1 kg = 75.0 kg 2.20 lb

Example Problems

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A bucket contains 4.65 L of water. How manygallons of water is that?

Unit plan: L qt gallon

Equalities: 1.06 qt = 1 L 1 gal = 4 qt

Set up Problem:

More examples

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Given: 4.65 L Needed: gallons

Plan: L qt gallon

Equalities: 1.06 qt = 1 L; 1 gal = 4 qt

Set Up Problem:

4.65 L x x 1.06 qt x 1 gal = 1.23 gal

1 L 4 qt

3 SF 3 SF exact 3 SF

Solution

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Density


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Density

• Compares the mass of an object to its volume.

• Is the mass of a substance divided by its volume.

Density expressionDensity = mass = g or g = g/cm3 volume mL cm3

Note: 1 mL = 1 cm3

Density

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Densities of Common Substances

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Osmium is a very dense metal.

What is its density in g/cm3 if 50.0 g of osmium has a volume of 2.22 cm3?

1) 2.25 g/cm3

2) 22.5 g/cm3

3) 111 g/cm3

Example

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Given: mass = 50.0 g volume = 22.2 cm3

Plan: Place the mass and volume of the osmium metalin the density expression.

D = mass = 50.0 g volume 2.22 cm3

calculator = 22.522522 g/cm3

final answer (2) = 22.5 g/cm3 (3 SF)

Solution

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Density

• How many mL of mercury are in a thermometer that contains 20.4 g of mercury?

• Have: mass

• Need: volume

• Use: density of mercury (d=13.6 g/mL)

found in Table 1.11

d = m/vol vol = m/d

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Sink or Float

• Ice floats in water because the density of ice is less than the density of water.

• Aluminum sinks because its density is greater than the density of water. Copyright © 2005 by Pearson Education, Inc.

Publishing as Benjamin Cummings

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Which diagram correctly represents the liquid layers in the cylinder? Karo (K) syrup (1.4 g/mL), vegetable (V) oil (0.91 g/mL,) water (W) (1.0 g/mL)

1 2 3

K

K

W

W

W

V

V

V

K

Think about it!

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Volume by Displacement

• A solid completely submerged in water displaces its own volume of water.

• The volume of the solid is calculated from the volume difference.

45.0 mL - 35.5 mL

= 9.5 mL = 9.5 cm3


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Density Using Volume Displacement

The density of the zinc object isthen calculated from its massand volume.

mass = 68.60 g = 7.2 g/cm3 volume 9.5 cm3


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Specific Gravity

• Ratio between the density of a substance and the density of water (= 1.00 g/ml)

Example:

Coconut oil has a density of 0.925 g/ml.

What is the specific gravity?

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Solution

Specific gravity of coconut oil:

Sp gr =

density of oil

density of water=

0.925 g/ml

1.00 g/ml= 0.925

no units!!

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More examples: Specific Gravity

What is the specific gravity of ice if 35.0 g of ice has a volume of 38.2 ml?

Density of ice = mass/volume

Specific gravity = density of ice/density of water

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Chapter 2.3 Temperature Conversion


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Temperature Scales

Temperature Scales


• are Fahrenheit, Celsius, and Kelvin.

• have reference points for the boiling and freezing points of water.

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A. What is the temperature of freezing water?

1) 0°F 2) 0°C 3) 0 K

B. What is the temperature of boiling water?

1) 100°F 2) 32°F 3) 373 K

C. How many Celsius units are between the boiling and freezing points of water?

1) 100 2) 180 3) 273

Think about it

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Temperature Conversion

Fahrenheit

Celsius

TF = 1.8(TC) + 32Tc =TF - 32

1.8

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Celsius - Fahrenheit

A person with hypothermia has abody temperature of 34.8°C. What is that temperature in °F?

TF = 1.8 TC + 32

TF = 1.8 (34.8°C) + 32° exact tenth's exact

= 62.6 + 32° = 94.6°F tenth’s Copyright © 2005 by Pearson Education, Inc.

Publishing as Benjamin Cummings

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The normal temperature of a chickadee is 105.8°F. What is that temperature on the Celsius scale?

1) 73.8°C

2) 58.8°C

3) 41.0°C


TC= (TF – 32°)/1.8

= (105.8°F - 32°)/1.8 = 41°C

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A pepperoni pizza is baked at 455°F. What temperature is needed on the Celsius scale?

1) 423°C

2) 235°C

3) 221°C

One more time

TC= (TF – 32°)/1.8

= (455°F - 32°)/1.8 = 235°C

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On a cold winter day, the temperature is –15°C.What is that temperature in °F?

1) 19°F

2) 59°F

3) 5°F

Learning Check

TF = 1.8(TC) + 32° = 1.8(-15°) + 32° = 5°F

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The Kelvin temperature scale• has 100 units between the freezing and boiling

points of water.

• is obtained by adding 273 to the Celsius temperature.

TK = TC + 273

• contains the lowest possible temperature, absolute zero (0 K).

0 K = –273°C

Kelvin Temperature Scale

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Fahrenheit

CelsiusKelvin

TF = 1.8(TC) + 32

Tc =TF - 32

1.8TK = TC +273

TC = TK - 273

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Temperatures

TABLE 2.5

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What is normal body temperature of 37.0°C in Kelvins?

1) 236 K

2) 310 K

3) 342 K

Calculate

TK = TC + 273 = 37.0°C + 273 = 310. K

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On the planet Mercury, the average night temperature is 13 K, and the average day temperature is 683 K.

1. What are these temperatures in Celsius degrees?

2. In Fahrenheit?TC = TK -273

TF = 1.8(TC) +32

1 problem solving using conversion factors copyright © 2005 by pearson education, inc. publishing...

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