Question

Calculate the carbon footprint (the emission of CO2 in kg per kWh of electricity generated) for a coal - fired power plant with an energy efficiency of 33%. The coal source contains 60% (weight/weight) carbon. The energy efficiency of the power plant is based on gross heat of combustion.

Answer 1

Ans. 1 kWh = 3.6 x 10³ J

# Calorific value of carbon = 3.25 x 10⁶ J/ kg   

# Step1: Given, efficiency of the power plant = 33%.

Let total amount of energy produced from combustion = X. Only 33% of X is converted into electricity (1 kWh).

So,

            33% of X = 3.6 x 10⁶ J

            Or, X = (3.6 x 10⁶ J) / 0.33

            Hence, X = 1.1 x 10⁷ J

  Therefore, total amount of heat to be produced to generate 1 kWh electricity = 1.1 x 10⁷ J

# Step 2: Amount of carbon required = Required heat / Calorific value of carbon

                                                = 1.0909 x 10⁷ J / (3.25 x 10⁶ J/ kg)

                                                = 3.38461 kg

                                                = 3384.6 g

# Step 3. Moles of C in 3.38 kg sample = Mass / Molar mass

                                                = 3384.6 g / (12 g/ mol)

                                                = 282.05 mol

# 1 mol C produces 1 mol CO_­2.

So,

            Moles of CO₂ produced = Required moles of C = 282.05 mol

            Mass of CO₂ produced = Moles x Molar mass

                                                = 282.05 mol x (44.0 g/mol)

                                                = 12410.2 g

                                                = 12.41 kg

# Therefore, carob footprint is “12.41 kg CO₂ per kWh”