Question

A coal-burning power plant generates electrical power at a rate of 750 megawatts (MW), or 7.50 × 108 J/s. The plant has an overall efficiency of 35% for the conversion of heat to electricity. Given this efficiency, how much heat energy in joules is produced in one year of operation?

Answer 1

Answer:

Heat Energy produced in one year of operation is 6.758 x 10¹⁶ J

Explanation:

Given;

rate of power generation, E = 7.50 × 10⁸ J/s

overall  efficiency, η = 35%

To determine the amount of heat energy in joules produced in one year of operation;

first: convert one year to seconds

Assume there are 365 days in one year,

365 days x 24 hours x 60 minutes x 60 seconds = 31536000 s

Heat Energy produced in one year of operation at given rate = (7.50 × 10⁸ J/s) x (31536000 s)

= 2.3652 x 10¹⁶ J

Heat Energy produced in one year of operation at given efficiency =

(2.3652 x 10¹⁶ J) / (0.35) = 6.758 x 10¹⁶ J

Therefore, Heat Energy produced in one year of operation is 6.758 x 10¹⁶ J

Answer 2

The amount of heat energy in Joules that is produced in one year of operation is  $6.83 \times 10^{16}$ Joules.

Given the following data:

• Power = 750 megawatts (MW) or $7.50 \times 10^8$ J/s.

• Overall efficiency = 35%

To find the amount of heat energy in Joules that is produced in one year of operation:

First of all, we would determine the time duration in seconds.

1 year = 365 days = 31536000 seconds

Next, we would determine the energy generated within this time frame or period.

Mathematically, energy is given by the formula:

$Energy = Power \times time$

Substituting the given parameters into the formula, we have;

$Energy = 7.50 \times 10^8 \times 31536000\\\\Energy = 2.39 \times 10^{16}\; Joules$

Heat energy at an overall efficiency of 35%:

$Heat\;energy = \frac{Energy\;produced}{Efficiency}\\\\Heat\;energy = \frac{2.39 \times 10^{16}}{0.35}$

Heat energy = $6.83 \times 10^{16}$ Joules

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