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Energy - Renewable and Non-renewable Energy

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Energy sources are classified as either renewable or non-renewable based on their replenishment rate. Non-renewable sources, such as fossil fuels (coal, oil, natural gas) and nuclear fuel (235U^{235}U), are depleted faster than they are formed.

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Renewable energy sources include solar, wind, hydroelectric, geothermal, and biomass. These are naturally replenished on a human timescale.

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Specific Energy is the amount of energy stored per unit mass of a fuel, measured in J⋅kg−1J \cdot kg^{-1}. It is calculated as Es=EmE_s = \frac{E}{m}.

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Energy Density is the amount of energy stored per unit volume of a fuel, measured in J⋅m−3J \cdot m^{-3}. It is calculated as Ed=EVE_d = \frac{E}{V}.

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Energy degradation: In every energy transformation, some energy is converted into thermal energy that is dissipated to the surroundings and is no longer available to do useful work. This relates to the second law of thermodynamics.

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Efficiency (η\eta) is the ratio of useful energy output to the total energy input. Since some energy is always 'lost' as heat, η<100%\eta < 100\%.

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Environmental impact: Non-renewable sources often release greenhouse gases like CO2CO_2, contributing to global warming, whereas most renewables have a lower carbon footprint but may have other ecological impacts (e.g., habitat disruption from dams).

📐Formulae

Efficiency (\eta)=Useful Energy OutputTotal Energy Input×100%\text{Efficiency (\eta)} = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\%

Specific Energy=EnergyMass\text{Specific Energy} = \frac{\text{Energy}}{\text{Mass}}

Energy Density=EnergyVolume\text{Energy Density} = \frac{\text{Energy}}{\text{Volume}}

P=EtP = \frac{E}{t}

Ep=mghE_p = mgh (Relevant for Hydroelectric Power)

Ek=12mv2E_k = \frac{1}{2}mv^2 (Relevant for Wind Power)

💡Examples

Problem 1:

A coal-fired power station burns 2000 kg2000 \text{ kg} of coal to produce 4.8×1010 J4.8 \times 10^{10} \text{ J} of thermal energy. Calculate the specific energy of the coal used.

Solution:

Specific Energy=Em\text{Specific Energy} = \frac{E}{m} Specific Energy=4.8×1010 J2000 kg\text{Specific Energy} = \frac{4.8 \times 10^{10} \text{ J}}{2000 \text{ kg}} Specific Energy=2.4×107 J kg−1\text{Specific Energy} = 2.4 \times 10^7 \text{ J kg}^{-1}

Explanation:

To find the specific energy, we divide the total energy released by the mass of the fuel consumed. The result shows how many Joules are available in every kilogram of coal.

Problem 2:

A natural gas power plant has a total energy input of 500 MJ500 \text{ MJ} and produces 200 MJ200 \text{ MJ} of useful electrical energy. Calculate the efficiency of the power plant.

Solution:

η=Useful Energy OutputTotal Energy Input×100%\eta = \frac{\text{Useful Energy Output}}{\text{Total Energy Input}} \times 100\% η=200 MJ500 MJ×100%\eta = \frac{200 \text{ MJ}}{500 \text{ MJ}} \times 100\% η=0.4×100%=40%\eta = 0.4 \times 100\% = 40\%

Explanation:

Efficiency is the percentage of input energy that is converted into the desired output. Here, 300 MJ300 \text{ MJ} of energy is wasted as heat, resulting in a 40%40\% efficiency rate.

Problem 3:

A hydroelectric dam uses water falling from a height of 50 m50 \text{ m}. If 1000 kg1000 \text{ kg} of water flows every second, calculate the theoretical power input available from the water. (Use g=9.8 m s−2g = 9.8 \text{ m s}^{-2})

Solution:

Ep=mghE_p = mgh Ep=1000×9.8×50E_p = 1000 \times 9.8 \times 50 Ep=490000 JE_p = 490000 \text{ J} P=Et=490000 J1 s=490000 W=490 kWP = \frac{E}{t} = \frac{490000 \text{ J}}{1 \text{ s}} = 490000 \text{ W} = 490 \text{ kW}

Explanation:

The gravitational potential energy of the water is converted into kinetic energy and then electrical energy. Since 1000 kg1000 \text{ kg} flows every second, the energy per second (power) is calculated using the potential energy formula.