Questions: A new coal-burning power plant can generate 2.1 gigawatts (billion watts) of power. Burning 1 kilogram of coal yields about 445 kilowatt-hours of energy. Complete parts (a), (b), and (c) below. a. How much energy, in kilowatt-hours, can the plant generate each month? The plant can generate 1,512 million kilowatt-hours of energy each month. (Type an integer.) b. How much coal, in kilograms, is needed by this power plant each month? The plant needs million kilograms of coal each month. (Round to one decimal place as needed.)

Solution Steps

First, we need to determine the total energy generated by the power plant in one month. Given that the power plant generates 2.1 gigawatts (GW) of power continuously, we can convert this to kilowatt-hours (kWh) over a month.

1 gigawatt (GW) = 1,000,000 kilowatts (kW)

So, 2.1 GW = 2,100,000 kW

There are 24 hours in a day and approximately 30 days in a month. Therefore, the total energy generated in a month is:

\[ \text{Energy per month} = 2,100,000 \, \text{kW} \times 24 \, \text{hours/day} \times 30 \, \text{days/month} \]

\[ \text{Energy per month} = 1,512,000,000 \, \text{kWh} \]

The problem states that the plant can generate 1,512 million kilowatt-hours of energy each month, which matches our calculation:

\[ 1,512,000,000 \, \text{kWh} = 1,512 \, \text{million kWh} \]

Next, we need to determine how much coal is required to generate this amount of energy. We know that burning 1 kilogram of coal yields about 445 kilowatt-hours of energy.

\[ \text{Coal needed} = \frac{\text{Total energy per month}}{\text{Energy per kilogram of coal}} \]

\[ \text{Coal needed} = \frac{1,512,000,000 \, \text{kWh}}{445 \, \text{kWh/kg}} \]

\[ \text{Coal needed} \approx 3,396,629.2135 \, \text{kg} \]

To convert this to millions of kilograms and round to one decimal place:

\[ \text{Coal needed} \approx 3.4 \, \text{million kg} \]

Final Answer