Question

An incandescent light bulb operated for two hours per day uses 12.2 kWh of energy per month. Burning of one kg of coal generates 2 kWh of electrical energy and releases 7 g of PM10. The reduction in PM10 emitted per month, if this incandescent bulb is replaced with a light emitting diode (LED) bulb which consumes 1/6th of energy, is ________ g (rounded off to one decimal place).

Hint:

To calculate the reduction in emissions, determine the energy saved and then calculate the amount of coal needed to generate that energy. Multiply the amount of coal by the PM10 emission factor.

Answer 1

Step 1: Calculate the energy consumed by the LED bulb.
The energy consumed by the incandescent bulb is given as 12.2 kWh per month. The LED bulb consumes 1/6th of this energy: \[ \text{Energy consumed by LED} = \frac{12.2 \, \text{kWh}}{6} = 2.0333 \, \text{kWh} \] Step 2: Calculate the energy saved by replacing the incandescent bulb with the LED bulb.
The energy saved per month is the difference in energy consumption between the incandescent bulb and the LED bulb: \[ \text{Energy saved} = 12.2 \, \text{kWh} - 2.0333 \, \text{kWh} = 10.1667 \, \text{kWh} \] Step 3: Calculate the amount of coal needed to generate the saved energy.
Burning 1 kg of coal generates 2 kWh of energy. Therefore, the amount of coal required to generate 10.1667 kWh is: \[ \text{Amount of coal} = \frac{10.1667 \, \text{kWh}}{2 \, \text{kWh/kg}} = 5.0833 \, \text{kg} \] Step 4: Calculate the reduction in PM10 emissions.
Burning 1 kg of coal releases 7 g of PM10. Therefore, the reduction in PM10 emissions for 5.0833 kg of coal is: \[ \text{PM10 reduction} = 5.0833 \, \text{kg} \times 7 \, \text{g/kg} = 35.5831 \, \text{g} \] Step 5: Round the result to one decimal place.
\[ \text{PM10 reduction} \approx 35.6 \, \text{g} \]