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1 Problem Solving using Conversion Factors Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings.

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Presentation on theme: "1 Problem Solving using Conversion Factors Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings."— Presentation transcript:

1 1 Problem Solving using Conversion Factors Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

2 2 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

3 3 Learning Check An injured person loses 0.30 pints of blood. How many milliliters of blood would that be? 1. Identify the given and needed units given in this problem. Given unit= _______ Needed unit = _______ pints milliliters

4 4 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 2 Unit 1 Given x Conversion= Needed unit factor unit contd # pints = # milliliters pints  milliliters

5 5 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

6 6 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 Needed unit factor unit 2.5 hr x 60 min = 150 min ( 2 SF ) 1 hr Setting up a Problem Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

7 7 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

8 8 A rattlesnake is 2.44 m long. How many centimeters long is the snake? Given Conversion Needed unit factor unit 2.44 m x 100 cm = 244 cm 1 m Solution

9 9 Often, two or more conversion factors are required to obtain the unit needed for the answer. Unit 1 Unit 2Unit 3 Additional conversion factors are placed in the setup to cancel each preceding unit Given unit x factor 1 x factor 2 = needed unit Unit 1 x Unit 2 x Unit 3 = Unit 3 Unit 1 Unit 2 Using Two or More Factors

10 10 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 10 3 min

11 11 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 = day 2 /min is not the unit needed Units don’t cancel properly. Check the Unit Cancellation

12 12 What is 165 lb in kg? STEP 1 Given 165 lb Need kg STEP 2 Plan STEP 3 Equalities/Factors 1 kg = 2.20 lb 2.20 lb and 1 kg 1 kg 2.20 lb STEP 4 Set Up Problem 165 lb x 1 kg = 75.0 kg 2.20 lb Example Problems

13 13 A bucket contains 4.65 L of water. How many gallons of water is that? Unit plan: L qt gallon Equalities:1.06 qt = 1 L 1 gal = 4 qt Set up Problem: More examples

14 14 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

15 15 Density Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

16 16 Density Compares the mass of an object to its volume. Is the mass of a substance divided by its volume. Density expression Density = mass = g or g = g/cm 3 volume mL cm 3 Note: 1 mL = 1 cm 3 Density

17 17 Densities of Common Substances

18 18 Osmium is a very dense metal. What is its density in g/cm 3 if 50.0 g of osmium has a volume of 2.22 cm 3 ? 1) 2.25 g/cm 3 2) 22.5 g/cm 3 3) 111 g/cm 3 Example

19 19 Given: mass = 50.0 g volume = 22.2 cm 3 Plan: Place the mass and volume of the osmium metal in the density expression. D = mass = 50.0 g volume2.22 cm 3 calculator = g/cm 3 final answer (2) = 22.5 g/cm 3 (3 SF) Solution

20 20 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

21 21 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

22 22 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) K K W W W V V V K Think about it!

23 23 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 mL mL = 9.5 mL = 9.5 cm 3 Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

24 24 Density Using Volume Displacement The density of the zinc object is then calculated from its mass and volume. mass = g = 7.2 g/cm 3 volume 9.5 cm 3 Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

25 25 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 g/ml. What is the specific gravity?

26 26 Solution Specific gravity of coconut oil: Sp gr = density of oil density of water = g/ml 1.00 g/ml =0.925 no units!!

27 27 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

28 28 Chapter 2.3 Temperature Conversion Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

29 29 Temperature Scales Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings are Fahrenheit, Celsius, and Kelvin. have reference points for the boiling and freezing points of water.

30 30 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) 1002) 1803) 273 Think about it

31 31 Temperature Conversion Fahrenheit Celsius T F = 1.8(T C ) + 32  T c = T F - 32  1.8

32 32 Celsius - Fahrenheit A person with hypothermia has a body temperature of 34.8°C. What is that temperature in °F? T F = 1.8 T C + 32  T F = 1.8 (34.8°C) + 32° exact tenth's exact = ° = 94.6°F tenth’s Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

33 33 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 Temperature Conversion T C = (T F – 32°)/1.8 = (105.8°F - 32°)/1.8 = 41°C

34 34 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 T C = (T F – 32°)/1.8 = (455°F - 32°)/1.8 = 235°C

35 35 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 T F = 1.8(T C ) + 32° = 1.8(-15°) + 32° = 5°F

36 36 The Kelvin temperature scale has 100 units between the freezing and boiling points of water. is obtained by adding 273 to the Celsius temperature. T K = T C contains the lowest possible temperature, absolute zero (0 K). 0 K = –273°C Kelvin Temperature Scale

37 37 Temperature Conversion Fahrenheit Celsius Kelvin T F = 1.8(T C ) + 32  T c = T F - 32  1.8 T K = T C +273 T C = T K - 273

38 38 Temperatures TABLE 2.5

39 39 What is normal body temperature of 37.0°C in Kelvins? 1) 236 K 2) 310 K 3)342 K Calculate T K = T C = 37.0°C = 310. K

40 40 Temperature Conversion 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? T C = T K -273 T F = 1.8(T C ) +32 


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