Comparing CO2 pollution using heat from incandescent light bulbs versus a gas heater

In terms of light produced from incandescent light bulbs versus compact fluorescent bulbs (CFLs), the CFLs use only one quarter of the electricity and produce only one quarter of the CO2 pollution as incandescent bulbs do.  However, since incandescents produce 90% heat, the efficiency of the bulbs as electric heaters in the winter should be compared to that of gas heaters as far as CO2 pollution is concerned.  Also, in the summer, the extra heat from incandescents may require AC usage to remove the heat.  We try to estimate the amount of CO2 produced between the two type of bulbs including the heating and AC effects, as well as comparing coal versus natural gas electricity sources.

In the winter, we first find the efficiency for heat and CO2 produced using the incandescent bulb as a heat source versus a gas heater.  First, assuming a natural gas plant electricity source, only 33% of its burned energy goes into electricity.  Then transmission may remove 10%, and the incandescent bulb gives off 10% as light.  However, if the light is absorbed in the room and does not travel out through a window, then it eventually produces heat in the objects or walls in the room.  So we count the heat output as 100% of the power rating for the incandescent bulb.  This leaves the efficiency of the gas plant heating by an incandescent bulb down to 30% of the inherent natural gas potential energy and its CO2 generation.  We compare this to a natural gas home heater which may be only 80% efficient due to vent leaking.  So to get the same amount of heat from both, we need to burn 80/30 = 2.7 more natural gas in the electricity plant than in the home gas heater, and generate 2.7 times as much CO2.  If the electricity plant is in a region where the main power is coal, which generates about 1.8 times the CO2 for the same electricity, the ratio of CO2 produced is increased to 2.67 x 1.8 = 4.8 times as much.

In winter, a CFL bulb which uses one quarter the power for the same amount of light, with the light being absorbed in the room, will generate one quarter of the heat.  If we desire to have the same heat as the incandescent bulb in the room, the rest of the heat of the incandescent bulb will need to be made up by the gas heater.  For a kiloWatt heat energy from an incandescent bulb set, the natural gas burned in the electric plant must have used 1.00/.30 = 3.33 kiloWatt of potential in the natural gas. The 0.25 kiloWatts of heat generated by the CFL set comes from 0.25 x 1.00 / .30 = 0.83 kiloWatts in the natural gas potential energy used electric plant.  To get the remaining 0.75 kiloWatts of heat from the gas heater requires 0.75 / 0.80 = 0.94 kiloWatts of natural gas potential energy.  The total of these for the CFL is (0.73 + 0.94) = 1.67 kiloWatts of natural gas potential energy.  The ratios of the natural gas potential energy for the incandescent versus the CFL bulbs to get a kiloWatt of heat energy and the same amount of light is 3.33/1.67 = 2.00 or twice as much.

In the summer, if it is hot enough that an AC is needed to cool from the incandescent light bulbs burning, we note that an AC is only 60% efficient in using electrical energy to remove heat.  So for a kiloWatt of incandescent bulbs, generating a kiloWatt of heat, we have to add 1.0 / 0.60 = 1.67 times as much electricity to cool from the bulb, or 1.67 kiloWatt.  So we run a total of 2.67 kiloWatts to get the equivalent of a kiloWatt of incandescent light that is heat neutral.  Compare this to the electricity required for an equivalent CFL that only uses 0.25 kiloWatt to start.  It generates 0.25 kiloWatts of heat.  To cool the room back to neutral, you need 0.25 kiloWatt / 0.60 = 0.42 kiloWatts of AC power.  Thus to generate CFL light which is heat neutral requires a total of 0.25 + 0.42 = 0.67 kiloWatt.  The ratio of power or CO2 pollution between incandescent and CFL light that is heat neutral in the summer is then 2.67 / 0.67 = 4.0 or four times as much.  I now realize that

1 x (1 + 1/.6)/.25 (1+1/.6) = 4, the ratio we started with for the power of the equivalent light bulbs.

About Dennis SILVERMAN

I am a retired Professor of Physics and Astronomy at U C Irvine. For two decades I have been active in learning about energy and the environment, and in reporting on those topics for a decade. For the last four years I have added science policy. Lately, I have been reporting on the Covid-19 pandemic of our times.
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