Skip Navigation


Household Energy Costs and Conservation 

Harvey F. Blanck

Department of Chemistry

January  2013

Austin Peay State University, 
Clarksville, Tennessee

(original 1979)


I.    Operating cost of electrical equipment


A.  Units and Cost 

A watt is a measure of the rate at which electrical energy is consumed.  Since power is the rate at which energy is consumed, a watt is a unit of electrical power.  Energy consumed will therefore be the rate at which it is consumed multiplied by the time the equipment is operating.  The energy consumed as reported on a monthly bill is in kwh (kilowatt hours).

To calculate energy consumed by a particular device multiply the kilowatts (watts x 1000) by the hours used.  Some devices will list watts while others will list amps and volts.  Watts equals amps times volts. All calculations that follow are based on a cost of 10 cents per kwh.

 B.   Definition of 'appliance' as used in this article

In this article the word 'appliance' will refer to all electrical devices such as a TV, toaster, stove, light, or food blender but not heating or cooling systems, which will be treated separately.


 II.   Heating costs for an electrically heated building

       The hourly BTU heat loss from a building can be calculated from the characteristics of the building and the interior and exterior temperatures or from the electrical energy consumed.  Either method may be used.  The latter is a more accurate way to determine the present heating costs, while the former is more useful in determining how operating costs may best be reduced.  As an example, suppose a house during the winter requires 14,000 BTU per hour when the outside temperature is 30 deg F and the inside temperature is 70 deg F.   How much does it cost to heat this house with electricity and how might energy costs be reduced?

A.  Houses heated electrically (except using heat pumps).

 1)   For electric resistive heaters one kilowatt hour consumed produces one kilowatt hour of heat.  (To convert from BTU's per hour to watts, multiply by 0.293 or from watts to BTU's per hour multiply by 3.41.)  To supply 14,000 BTU's per hour requires an average of 4100 watts continuously which costs 41 cents per hour or $9.84 per day or almost $300 per 30 day month using electrical heating methods other than heat pumps.

 2)   Since rate of energy loss depends upon the temperature difference, a 'heating degree-day' has been devised as a measure of the average temperature difference between inside and outside temperatures. Degree-days per winter may be determined by taking the average temperature difference between the inside and outside during the winter for each day (or approximately by using the average of the daily high and low temperature) and summing this for the entire winter. Because individual home owners use different thermostat settings, if degree-days are to be used for comparison of heating demands in various locations and for comparison of one winter season with another, it is necessary to use a fixed inside reference temperature. The reference temperature currently in use is 65 deg F.  Using this reference temperature the average winter in the Clarksville, Tennessee area produces about 3600 heating degree-days from November through March.  For example, if the average temperature in January for this area were 30 deg F,  there would be [(65 - 30) degrees x 31 days] or 1085 heating degree-day units.  If the inside temperature is actually kept at 70 deg F, then the actual relative heat loss will be the equivalent of 155 degree-days (5 degrees x 31 days) higher for a typical January.  For the entire winter this would be the equivalent of 750 degree-days (5 degrees x 150 days) higher.  Therefore, for an inside temperature of 70 deg F the actual equivalent heating degree-days per winter would be about 4350 rather than 3600.

 For the example house above the $300 for the month is for 1200 actual heating degree-days [(70 - 30) x 30].  The cost per winter is therefore about 3.6 times this, which is about $1080.

 Note: As will be seen shortly, heat is actually derived not only from the heating system but also from other sources including electrical appliances.  Therefore the 3600 heating degree-day units may be thought of as roughly representing the heating demand placed upon the heat generating devices.  Thus the $1080 estimate includes the cost of operating electrical appliances and an electrical heating system.  (Heat pump systems will be covered separately later.)

 3)   It is very important to realize that almost all the energy used to operate electrical appliances except a clothes dryer and possibly a hot water heater ends up as heat in the house.   This is a consequence of the First Law of Thermodynamics which states that energy can be converted from one form to another but it can not be destroyed.  (We often say energy is 'consumed' when it would be more accurate to say energy is 'converted'.)   For example, operating a 70 watt TV contributes 70 watts of heat to the house.  If one used about 59 of these 70 watt TV's, they would supply the 4100 watts to heat the house mentioned above.  Or you could turn on 41 lamps that had 100 watt light bulbs.  Or you could use some combination of lights and TV's to supply all or part of the 4100 watts.  The consequence of this for the example house is that the total energy produced by an electric heater plus the energy 'consumed' by appliances must add up to 4100 watts.  Reading by a small light with the rest of the house dark during the winter simply means that the electric furnace comes on more often than had all the lights and TV's been on in the house.  (In the summer, however, you pay more by having numerous lights and TV's on in an air conditioned house.)

       Example:  If the appliance operating cost is $100 per month, then the electric furnace cost for our example house will be $300 - $100 = $200.    If by using appliances frugally, the appliance operating cost is reduced to $25, then the electric furnace cost will be $300 - $25 = $275.   The total energy cost is the same --- $300.  (It is easy to become discouraged about conserving energy if you do not go about it correctly.)

B.   Houses heated electrically using heat pumps

 1)   Heat pumps use electrical energy to transport thermal energy (heat) from outside the house to the inside.  It is more efficient to use the electrical energy to transport thermal energy than to change the electric energy directly into heat as is done with resistive electric furnaces.   Heat pump operation may be listed as BTU input per hour rather than watts.  The BTU output from the heat pump depends upon the temperature difference between the inside and outside temperatures.  The greater the temperature difference the more difficult it is to transport heat.   The ratio of the BTU output over the BTU input is called the coefficient of performance, COP.  Heat pump COP's are listed at 47 deg F and 17 deg F.   For example, a heat pump COP at 47 deg F might be 2.8 but only 1.9 at 17 deg F.   A COP of 2.8 means that for every BTU put in, 2.8 BTU's are produced. 

 2)   To derive all the heating needs from a heat pump during the coldest winter night might require the use of a heat pump four times as large as one that would be adequate at 30 deg F.  To reduce heat pump purchase price, units may be designed to use the heat pump down to 30 deg F and then use auxiliary resistive electric heaters to help out the heat pump at lower temperatures.  These auxiliary heaters like all resistive heaters produce one BTU of heat for each BTU of energy input and are therefore more expensive to operate.  (Note: Auxiliary heaters may also come on if the thermostat setting is suddenly increased significantly.  The auxiliary heaters are used in order to help get the temperature of the house up to the new setting rapidly for occupant comfort at the expense of energy conservation.

3)   If the example house used a heat pump with an average COP of 2, the heating costs for operating the heat pump would be half that of operating a resistive heating electric furnace.  Appliance operation will reduce the heat pump operation time but now only half the cost of operating the appliances may be subtracted from the cost of operating the more efficient heat pump.   If the example house used $80 for appliance operation, the heat pump operating cost would be reduced $40.  The cost to heat the house with a heat pump with no contribution from appliances would be $300/2 = $150.   $80 worth of energy from operating appliances would reduce the heat pump cost from $150 to $110 (i.e.150- 40).  The total electric bill would then be heat pump plus appliance operating cost or $110 + $80 = $190.   If a heat pump is used, using appliances conservatively will reduce electrical energy cost.

 Note:   For houses heated with gas much of the above discussion applies.  Electrical appliance operation will reduce the amount of gas used to operate the gas furnace.  Cost of gas and electricity must be compared to see how much, if any, money is saved by conservative use of electrical appliances.

 III.    Sample calculations of approximate heat loss

 A.  Formula:  BTU loss per hour  = (area in square feet / R value)  x  temp difference in deg F

 B.   Calculations are for a house which is 24' x 62.5' (1500 sq ft) with 8' ceilings having wallboard ceiling and interior walls, polystyrene sheathing (R3), brick exterior, carpeted floors, and 8% of the wall area window or door glass.  All glass is double pane or single pane with storm windows.   Inside temperature 70 deg F and outside temperature including attic 30 deg F and basement or crawl space 50 deg F.   R is the inherent resistance to heat flow.  R values are additive.  For example, 3 inches of fiberglass has an R of 10 and 6 inches an R of 20.

 Note: Single pane glass has such poor insulating characteristics, that only double pane windows or single pane windows plus storm windows are considered in the examples.

 C.   Calculated heat loss for the above house with no fiberglass (or equivalent) insulation.  (Percentage is the % of BTU lost by that particular part of the house.)



house component





BTU loss

per hour




1500 sq ft






(excluding windows)

1557 sq ft







1500 sq ft






(double or storm)

140 sq ft













D. Calculated heat loss with fiberglass (or equivalent) insulation.



house component





BTU loss
per hour




1500 sq ft

6 inches





(excluding window)

1557 sq ft

3 inches






1500 sq ft

3 inches





(double or storm)

140 sq ft












 Note:  For this insulated house the BTU loss per hour is about 14,000 since heat is lost by infiltration of cold air and by opening exterior doors briefly which adds 10 - 15 %.


IV.    Air conditioning costs

 Air conditioning costs can be calculated in much the same way as heating costs.  The insulated house used as an example would have a heat flow inward of 7000 BTU per hour if all the temperature differences were one half those in winter.  That is: 70 deg F inside, 90 deg F outside and 80 deg F basement or crawl space.  Since a heat pump and an air conditioner operate on exactly the same principles, the COP of an air conditioner might be 2.5.  Thus a BTU input of 7000/ 2.5  =  2800  or 820 watts would be required to pump the thermal energy out that comes in through the walls, ceiling, floor, and windows.  In addition all energy from appliances must be pumped out.  Also, air conditioners remove water vapor (i.e. lower humidity) and since condensation of water from the air releases heat, that heat must be pumped out as well.  

 Note:  Air conditioners are rated by their energy efficiency ratio (EER) which is the BTU's per hour pumped out divided by the watts input (a weird apples and oranges kind of thing).   A COP of 1.5 has an EER of 5.1 and a COP of 3.0 has an EER of 10.2 thus energy cost of operation is halved if EER is doubled.   EER's are measured at 95 deg F.  Air conditioners are also rated at 82 deg F.  This is called the Seasonal EER or SEER.  Since the temperature difference is smaller, the SEER will be larger than the EER.   Since our summers can be downright hot, the EER is probably the more important of the two.


V.     Hot water costs

 A.  Cost per gallon of bath or shower water

       The cost per gallon of warm water used depends on a temperature difference.  If the cold water input is 60 deg F and the shower temperature is 105 deg F, the cost depends on the difference in temperature which in this case is 45 deg F.  (It does not matter how hot the hot water is provided it exceeds 105 deg F.)   One gallon of water weighs 8.34 pounds and requires 8.34 BTU to raise the temperature one degree.  In this case, the energy required to heat one gallon of water is 8.34 x  45 = 375 BTU which is 0.110 kwh at a cost of just a bit over one cent per gallon.

 B.   Bathing costs

 1)   A very modest tub bath is about 8 gallons or about 8 cents per bath.

 2)   A shower at a modest rate consumes about one gallon per minute.  An 8 minute shower costs the same as a meager bath.   

 C.   Laundry and dishwashing hot water costs are quite variable but can be calculated in a similar fashion.

 D.  Other hot water costs include loss of heat from hot water left in the water line after the hot water tap is shut off depending on the location of the water line and the season of the year.  The hot water tank also continuously loses heat to its surroundings.  (There are alternatives to the use of a hot water tank.)

 VI.    Reducing energy consumption and cost

 A.  Eating plain bread rather than toast each morning might save about one dollar a year in toaster operation but not much energy or money is saved.  A toaster uses quite a bit of energy per unit time, but the total operating time per year is low.   Giving up toast is an ineffective way to conserve energy and save money.

 B.   It has been stated previously that electrical equipment operation is less costly during the winter months because heat from these devices reduces the time a furnace or heat pump is operating.  On the other hand, appliance operation is more costly during the summer if an air conditioner is operating, because heat generated by the appliance must be pumped out.  For example, a 100 watt light requires about 150 watts of electricity in an air conditioned house --- 100 watts to operate the light and 50 watts used by the air conditioner to pump the heat produced by the lamp outside the house if the air conditioner COP is 2.

C.   The rate of heat transport through a wall depends on the R value of the wall and the temperature difference between the inside and outside.  The outside temperature can not be readily controlled but the inside temperature can be.  If the temperature difference is decreased by 10% then energy transport through the wall is reduced 10% and it takes 10% less energy and money for heating and air conditioning.  Consider the following two examples. 

 1)   Winter:  If the inside temperature is 70 deg F and the outside temperature is 30 deg F, then lowering the inside temperature to 69 deg F will reduce energy consumption by 2.5%.  Had the inside temperature been lowered 40 deg (70-30=40) then 100% of the energy to heat the house would have been saved. Lowering the temperature 1 degree therefore saves one 40th of the energy.   ( 1/40 ) x 100  =  2.5 %.  Setting the temperature 4 degrees lower would save 10%.  Raising the temperature to 74 deg F would cost 10% more.

 2)   Summer: If the inside temperature is 70 deg F and the outside temperature is 90 deg F, then raising the inside temperature to 71 deg F will reduce energy consumption ( 1/20 ) x 100  =  5 %.  Setting the temperature 4 degrees higher would save 20%.  Lowering the temperature would cost 5% more for each degree.

 D.  Adding insulation to ceilings and floors is often a relatively inexpensive way to reduce energy loss.  R values are additive so doubling the amount of fiberglass (or equivalent) insulation will nearly double the R value of the ceiling.  For example, increasing the ceiling insulation of our example house from 6 inches of fiberglass to 12 inches would reduce the energy consumption about 11 %.  Often floors have little or inadequate insulation.  Increasing the floor fiberglass insulation from 3 inches to 6 inches can reduce the winter energy costs about 8 %.  (Caution:  Crawl spaces and hence water lines can freeze in the winter if outside air access is not controlled properly.)

 E.   Some homes have uninsulated garage doors in the basement or attached garages.  Uninsulated doors do keep the wind, rain, snow and your neighbor's cat out. Kits are available to insulate them.  In addition, be sure that garage walls and ceiling that are shared with heated interior living space are well insulated. 

 F.   Reducing the temperature of hot water tanks will save energy.  (To prevent accidental burns from hot water a temperature of 125 deg F is recommended by some experts.)   Additional insulation can be added to hot water tanks.  Seek instructions on how to do this.

 G.   Other energy saving tactics can be used. For example, controlling sunlight access to the house using trees, curtains, and blinds can significantly alter energy demands.  During much of the summer no air conditioning is needed at night if windows are open.  Removing the heat from a 100 watt bulb or from the refrigerator using an open window will often work just as well and much more efficiently than an air conditioner.  Adequate attic air circulation can significantly lower the attic temperature during the summer and reduce air conditioning energy usage.


VII.  Comparison of R values for various materials 




R value per inch

Thickness for R = 10


polystyrene foam


1.6 inches




2.5 inches


fiberglass or rock wool


3 inches




3 inches


sugar cane fiberboard (black)


3.5 inches




4 inches


wood (across grain)

0.9 - 1.5

7 - 11 inches


gypsum wallboard


10 inches


concrete block

0.3 - 0.5

2 - 3 feet



0.15 - 0.3

3 - 6 feet




4 feet



0.1 - 0.15

6 - 10 feet




14 feet




1,200 feet