Activity 1.3.3 Thermodynamics Answer Key



[pic]

Activity 1.3.3 Thermodynamics Answer Key

Introduction

Think back to the last time someone complained about a door being left open. What did you notice about the temperature within the room as a result of the open door? In Activity 1.3.3 you will investigate the effects of work, thermal energy, and energy on a system, as in the case of the room with the door left open.

Procedure

Answer the following questions as your teacher discusses the Introduction to Thermodynamics presentation.

Define thermodynamics.

The study of the effects of work, thermal energy, and energy on a system

List three examples of a thermodynamic system.

a. Nuclear power

b. Electronic heat sink

c. Rocket launch

Define thermal energy.

Kinetic energy in transit from one object to another due to temperature difference

Define temperature.

The average kinetic energy of particles in an object

|Scale |Freezing point of water |Boiling point of water |

|Celsius |0°C |100°C |

|Fahrenheit |32°F |212°F |

|Kelvin |273K |373K |

Fill in the table below with the correct scale and unit.

Define absolute zero.

All kinetic energy is removed - 0 K

Define thermal equilibrium.

Touching objects within a system reach the same temperature

Define the Zeroth Law of Thermodynamics.

If two systems are separately found to be in thermal equilibrium with a third system, the first two systems are in thermal equilibrium with each other.

Define the First Law of Thermodynamics.

Thermal energy can change form and location, but it cannot be created or destroyed.

List two ways thermal energy can be increased in a system.

d. Adding thermal energy

e. Performing work on the system

Define the Second Law of Thermodynamics.

Thermal energy flows from hot to cold.

Define entropy.

The measure of how evenly distributed heat is within a system

Define convection.

The transfer of thermal energy by movement of fluid (liquid or gas)

List two examples of convection.

f. Weather

g. Boiler systems

Define conduction.

The transfer of thermal energy within an object or between objects from molecule to molecule

List two examples of conduction.

h. Metal spoon

i. Heat through a wall

Conduction Equations:

[pic] [pic] [pic] [pic]

Define the following variables.

Q = Energy transferred

m = Mass of material absorbing or releasing energy

c = Specific heat capacity of a material (J/kg °C)

P = Rate of energy transfer

Δ t = Change in time

k = Thermal conductivity

A = Area of thermal conductivity

L = Thickness of material

Δ T = Change in Temperature

(18–22) A 1 kg piece of aluminum metal at 90 °C is placed in 4 liters (= 4 kg) of water at 25 °C. Determine the final temperature (Tf).

List all known values.

mAl = 1.00 kg

mwater = 4.00 kg

TAl = 90.0 °C

Twater = 25.0 °C

Cp-Al = 900. J/kg °C

Cp-water = 4184 J/kg °C

List all unknown values.

Tfinal

Select equations.

[pic] [pic]

Apply known values.

[pic]

[pic]

Solve.

[pic]

(23–27) The top of a 3/16 inch thick acrylic testing box is covered with an unknown ½ inch insulation material (black). The dimensions of box are 10 inch x 10 inch on each side. The sides and are wrapped with a control insulting material (blue) so that the almost all heat loss is through the 10 inch x 10 inch unknown insulating material on the top of the box (assume no heat is lost from the bottom or sides of the box). Determine the thermal conductivity for the insulating material if a 25 W bulb is used to heat the box. You may assume the only heat loss from the box occurs through the 10 x 10 test area. The bulb maintains the inside temperature at 10 ºC higher than the outside temperature.

[pic][pic]

List all known values.

A = 10.0 in. x 10.0 in. = 0.254 m x 0.254 m = 0.064516 m2

ΔT = 10.0 °C

P = 25.0 W

L = 0.50 in. = 0.0127 m

List all unknown values.

k = Thermal conductivity

Select equations.

[pic]

Apply known values.

[pic]

Solve.

[pic]

Define U-value.

Measure of a material’s ability to conduct heat

Define R-value.

Measure of a material’s ability to resist heat movement

(30–31) Use the provided R-value chart and illustration below to calculate the R-value of the wall cavity and the R-value at the stud location.

[pic]

Wall cavity R-value

5/8 drywall = 0.56

Fiberglass batt = 11.00 (3.5 in. x 3.142)

Extruded polystyrene = 4.00

Wood bevel lap siding = 0.80

0.56 + 11.00 + 4.00 + 0.80 = R16.36

R-value at stud location

5/8 drywall = 0.56

2 x4 stud = 4.38

Extruded polystyrene = 4.00

Wood bevel lap siding = 0.80

0.56 + 4.38 + 4.00 + 0.80 = R9.74

Define radiation.

The process by which energy is transmitted through a medium, including empty space, as electromagnetic waves

List two examples of radiation.

j. Microwave oven

k. Heat transfer from the human body

Define Stefan’s Law.

All objects lose and gain thermal energy by electromagnetic radiation.

(35–39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 58 °F. The student’s skin temperature is 91.4 °F. Determine the net energy transfer from the student’s body during the 20 min ride to school due to electromagnetic radiation. Note: Skin emissivity is 0.90, and the surface area of the student is 1.50 m2.

List all known values.

A = 1.50 m2

e = 0.90

ϭ = 5.6696 x 10-8 w/ m2 k 4

t = 20.00 min = 1200. s

Bus temperature = 58.0 °F = 287.60 K

Skin temperature = 91.4 °F = 306.15 K

List all unknown values.

P = Rate of energy transfer

Q = Energy transfer

Select equations.

[pic] [pic]

Apply known values to equations.

[pic]

Solve for (T4 –T4)

287.604 K = 6,841,566,093.34 K

306.154 K = 8,784,904,710.59 K

8,784,904,710.59 K - 6,841,566,093.34 K = 1,943,338,617.25 K

Pnet = (5.6696 x 10-8 w/ m2 k 4) x (1.50 m2) x (0.90) x (1,943,338,617.25 K)

= 148.74 W

[pic]

Q = 148.74 W x 1200. s

Solve.

Q = 178,000 J

Define geothermal energy.

Energy generated from the thermal energy stored beneath the Earth’s surface

|R-Value Chart |

|Construction Material |R-Value |

|½ in. Drywall |0.45 |

|5/8 in. Drywall |0.56 |

|Particle Board – ½ in. |0.63 |

|Particle Board – ¾ in. |0.94 |

|Fiberboard ½ in. |1.32 |

|Extruded Polystyrene 1 in. |4.00 |

|Extruded Polystyrene 1 ½ in. |6.00 |

|Foil-faced Polyisocyanurate 1 in. |7.20 |

|2 x 4 |4.38 |

|2 x 6 |6.88 |

|Hardwood |0.90 |

|Masonry Systems |R-Value |

|Brick 4 in. common |0.80 |

|Brick 4 in. face |0.44 |

|Concrete Block – Normal wt. 12 in. empty core |1.23 |

|Concrete Block – Light wt. 12 in. empty core |2.60–2.30 |

|Cement Mortar |0.20 |

|Sand and Gravel |0.60 |

|Stucco |0.20 |

|Roofing |R-Value |

|Asphalt Roll |0.15 |

|Asphalt Shingle |0.44 |

|Slate |0.05 |

|Wood |0.94 |

|Siding |R-Value |

|Wood Shingles |0.87 |

|Wood Drop |0.79 |

|Wood Bevel Lapped |0.80 |

|Aluminum/Steel – Hollow |0.61 |

|Aluminum/Steel – with 3/8 in. Backer |1.82 |

|Insulation |R-Value per in. |

|Fiberglass Batt |3.142 |

|Blankets – Rock Wool |3.0–3.8 |

|Loose Fill – Cellulose |2.8–3.7 |

|Loose Fill – Fiberglass 0.7 lb/cu.ft |2.2–4.0 |

|Loose Fill – Rock Wool |3.1 |

|Loose Fill – Vermiculite |2.2 |

|Extruded Polystyrene |4.00 |

[pic]

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download