Homework 3 Answers

Chapter 3. Question

1. Energy conservation means using as little energy as possible to
accomplish a given task.

Conservation of energy is a universal law of nature which
states that energy is never created or destroyed, merely changed from one
form into another. 

An example of the former is to turn off the lights when you leave the room
and don't need them. An example of the latter is when a falling object
converts potential energy into kinetic energy (or vice versa for a rising
object).

Chapter 3. Problems

2. According to Hinrichs, a gallon of gasoline has the heating value of
125,000 Btu. So, 20 gallons would be worth 2.5 million Btu. 

On the other hand, 90 pounds of coal is worth 1 million Btu, so 2.5 million
Btu = 2.5 x 90 = 225 pounds of coal. That is a lot!

4. 2 tons of coal is 4,000 pounds. Since 90 pounds of coal is worth 1 million
Btu (energywise, Table 3.4), 4,000 pounds is worth (4000/90)million Btu =
44.4 million Btu. 

Converting to kWh 1 kWh = 3413 Btu so 44.4 million Btu / (3413Btu/kWh) =
13,022 kWh.

The plant generates 6,000 kWh of electricity from the coal, so its efficiency
is 6000/13022 = 46%, which is pretty good!

Chapter 4. Questions

4. A body cannot contain heat, rather a body contains energy. This energy
can be transferred to another body and we call the energy that is
transfered "heat". Heat can be converted into other forms of energy.
Furthermore, a body cannot remain "hot" indefinitely (relative to its
surroundings), because energy is always radiated or otherwise transfered
to the surroundings, until all temperatures are the same.

5. From the first law of thermodynamics, which was not covered in time for
this problem set, we learn that work can be converted into internal energy
(or heat). A great example is friction, another is compressing a gas. When
you compress gas, you do work on it, and its temperature rises because the
internal energy is increased.

6.  Again, not covered in time for this problem set. Sketch should show
energy transfer by direct heat radiation (radiant or electromagnetic
energy), heat transfer by contact (e.g. a pot of boiling water on the
stove) and hot air rising from the stove (convection).

Chapter 4. Problems.

7. 240 gallons of water heated 23 degrees C or 41.5 degrees F.

Use Btu formula for energy. 1 Btu is the energy required to heat 1 pound 
of water by 1 degree F, or the heat capacity "c" = 1 Btu/pound/deg F

240 gallons = 1,920 pounds (8 pounds/gallon)

Total Btu required = mc delta T = 1920 lbs * 41.5 degrees F = 79,680 Btu.  
c = the heat capacity, which is 1 Btu/degree F/pound

Convert to kWh: 1 kWh = 3413 Btu, we need 79680/3413 = 23.3 kWh. 

8. At $0.08 /kWh this will cost 23.3 kWh*0.08/kWh = $1.87


9. In a hot water heater, 40 gallons heated from 60 to 120 F.  

Use Btu formula for total energy: 

40 gallons*8 pounds/gallon = 320 pounds.

320 pounds*60 degrees F = 19200 Btu required to heat the water

Convert to kWh: 19200 Btu/3413 (Btu/kWh) = 5.6 kWh. 

If we have a 4 kW immersion heater, this means that the heater 
supplies energy at a rate of 4 kWh per hour.

Since total = rate*time, time = total/rate.

Time to heat = 5.6 kWh /4 kWh per hour = 1.4 hour.