ch1410 lecture #3 tro ch2/3conversions and energy...

For most conversion problems, we are given a quantity in some units and asked to convert the quantity to another unit. These calculations take

the following form:

Unit-Conversion Methodology

information given x conversion factor(s) = information sought

given unit(s) x = desired unit(s)desired unit(s)given unit(s)

Unit Conversion Methodology

Step 1-Write the given measurement with its units. Identify the target units.

Step 2-Identify possible conversion factors.

Step 3-Multiply the given measurement by a conversion factor which allows you to cancel undesired units.

Step 4-Cancel units.

Step 5-Multiply by additional conversion factors if needed.

Step 6-Perform the actual calculation.

Unit-Conversion Methodology

Conversion factors are constructed from any two quantities known to be equivalent.

2.54 cm = 1.00 in

2.54 cm2.54 cm

1.00 in2.54 cm=

2.54 cm 1.00 in=

1.00 in 1.00 in

1.00 in2.54 cm

1 =2.54 cm

= 11.00 in

The Solution Map

A solution map is a visual outline that shows the strategic route required to solve a problem.

cm in1.00 in 2.54 cm

2.54 cm 1.00 in

in cm

The solution map for converting from centimeters to inches is:

The solution map for converting from inches to centimeters is:

Unit Conversion Method

How many inches are in 19.4 cm?

cm in

1.00 in2.54 cm

19.4 cm x = 7.63779528 = 7.64 in1.00 in2.54 cm

12.0 in

1.00 ft1.00 in2.54 cm

cm in ft

Unit Conversion Method

1.00 in2.54 cm 12.0 in

1.00 ft

194 cm x x = 6.3648 = 6.36 ft

How many feet are in 194 cm?

Factor-Label Method

How many kilometers are in 1 mile ? 1.609 km = 1.000 mi

0.6214 mi = 1.000 kmHow many miles are in 1 kilometer ?

How many kilometers are in 26.22 mile ?

1.609 km 1.000 mi

1.000 mi 1.609 km conversion factors

1.000 km 0.6214 mi

0.6214 mi 1.000 km conversion factors

26.22 mi X 1.609 km1.000 mi = 42.19 km

26.22 mi X 1.000 km0.6214 mi = 42.19 km

Conversion with Units Raised to a Power

We can do the same thing in fractional form:

We cube both sides to obtain the proper conversion factor:

(2.54 cm)3 = (1.00 in.)3 (2.543)cm3 = (1.00)3 in.316.387 cm3 = 1.00 in.3

=(1.00 in.)3

(2.54 cm)3 1.00 in3

2.54 cm3 =1.00 in.3

16.387 cm3

1.00 ft3 x x x = 28,317 mL

ft3 2.83 x104 mL

Unit Conversion Method

How many mL are in 1.00 ft3?

cm3 mLin3ft3

1728 in3 1 ft3

(12 in)3 1 ft3

(2.54 cm)3 1.00 in3

1.00 mL 1.00 cm3

16.387 cm3 1.00 in3

1.00 mL 1.00 cm3

8.00 cm3 x = 48.72 g

48.7 g

Density as a Conversion Factor

For a solid substance with a density of 6.09 g/cm3, what is the mass of 8.00 cm3 of the substance?

cm3 g

6.09 g 1.00 cm3

6.09 g 1.00 cm3

g cm3 mL

Density as a Conversion Factor

For a liquid substance with a density of 1.32 g/cm3, what is the volume of 68.4 g of the substance?

1.00 mL 1.00 cm3

1.00 cm3 1.32 g

68.4 g x x = 51.818182 mL

51.8 mL

1.00 mL 1.00 cm3

1.00 cm3 1.32 g

Energy

Energy is the capacity to do work

The law of conservation of energy states that energy is neither created nor destroyed.

The total amount of energy in the universe is constant.

Energy can be changed from one form to another.

Energy can be transferred from one object to another.

Units of Energy The SI unit of energy is the joule (J)

A second unit of energy is the calorie (cal), the amount of energy required to raise the temperature

of 1 g of water by 1 degree Celsius.

The nutritional Calorie (Cal) is equivalent to 1000 little c calories.

Energy Conversion Factors

Cal cal J

Conversion of Energy Units

A candy bar contains 225 Cal of nutritional energy. How many joules does it contain?

1000 cal 1 Cal

4.184 J 1 cal

225 Cal x x = 9.41 x105 J 1000 cal 1 Cal

4.184 J 1 cal

Temperature: Random Motion of Molecules and Atoms

The temperature of a substance is a measure of its thermal energy.

Heat, which has units of energy, is the transfer or exchange of thermal energy caused by a

temperature difference.

Comparison of Fahrenheit, Celsius, & Kelvin Scales

180 ºF 100 ºC 100 K

180/100 = 1.8 ºF 100/180 = 5/9

Converting between Temperature Scales

ºF = 1.8(ºC) +32

Temperature Changes: Heat Capacity

Heat capacity: The quantity of heat (usually in joules) required to change the temperature of a

given amount of a substance (or a specific object) by 1 °C. (J/°C)

Specific heat capacity (or the specific heat) The quantity of heat (usually in J/g) required to

change the temperature of one gram of a substance by 1 °C. (J/g °C).

Quantifying Heat Energy

We can calculate the quantity of heat absorbed by an object if we know the mass, the specific heat (for the particular type of matter), and the temperature change of the object.

J = g x x ºCJgºC

Heat = (mass) x (specific heat capacity) x (temp. change)

q = (m) x (C ) x (ΔT)

qm x C = ΔT

? If q and m are constant, what

happens to ΔT as C increases ?

Specific Heat Capacities of Common Substances

Specific Heat of Water

The rather high specific heat of water allows water to absorb a large quantity of heat energy without a large increase in its temperature.

Without water, the Earth’s temperature would be about the same as the moon’s temperature on the side that is facing the sun (average 107 °C or 225 °F).

Water is commonly used as a coolant because it can absorb a large quantity of heat and removes it from important mechanical parts to keep them from overheating or even melting.

4.18 J/(g·°C)

Relating Heat Energy to Temperature Changes

How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 29.9 °C? (The specific heat capacity of gallium is 0.372 J/g°C.)

q (J)= m (g) x C ( ) x ∆T (ºC)JgºC

C, m, ∆T q

q = (m)(C)(∆T)

How much heat must 2.5 g of gallium absorb from your hand to raise its temperature from 25.0 °C to 29.9 °C? (The specific heat capacity of gallium is 0.372 J/g °C.)

C, m, ∆T q

q = (m)(C)(∆T)

q = (m)(C)(∆T) = 2.5 g x 0.372 J/gºC x 4.9 ºC = 4.557 J = 4.6 J

C = 0.372 J/gºCm = 2.5 g∆T = Tf - Ti = 29.9 ºC - 25.5 ºC = 4.9 ºC

1. How many Joules of energy are needed to raise the temperature of __ g of a substance __ºC if its specific heat capacity is __ J/gºC ?2. How many grams of a substance with a specific heat capacity of __ J/gºC are present if __ Joules of energy raise the temperature of the substance __ºC?3. What is the specific heat capacity of a substance if __ Joules of energy raise the temperature of __ g of the substance __ºC?

4. What is the temperature change of __ g of a substance with a specific heat capacity of __ J/gºC a substance if __ Joules of energy are added?

C, m, ∆T q

q = (m) (C) (∆T)