Environmental Management and Sustainable Development · Energy and Environment · 2 marks · Numerical answer
An incandescent bulb used2 hours/day consumes12.2 kWh/month. Coal electricity yield is 2 kWh/kg and PM₁₀ emission is 7 g/kg coal. An LED uses one-sixth as much electricity. Find avoided monthly PM₁₀ in grams, to one decimal.
Energy saved is $12.2(1-1/6)=10.16667$ kWh/month. At2 kWh/kg, this avoids $10.16667/2=5.08333$ kg coal/month. Multiplying by 7 g PM₁₀/kg gives $35.58333$ g/month, rounded **35.6 g**. The usage schedule is already reflected in the supplied monthly consumption.
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