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Energy - Energy Sources and Resources

Grade 9IB

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

🔑Concepts

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Energy is defined as the capacity to do work, measured in Joules (JJ). The Law of Conservation of Energy states that energy cannot be created or destroyed, only transformed from one form to another.

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Renewable energy resources are those that are replenished naturally over short periods. Examples include solar, wind, hydroelectric, geothermal, tidal, and biomass.

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Non-renewable energy resources exist in finite amounts and take millions of years to form. These include fossil fuels (coal, oil, natural gas) and nuclear fuels like Uranium-235.

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Energy Density is the energy stored per unit volume (J m−3J \, m^{-3}), while Specific Energy is the energy stored per unit mass (J kg−1J \, kg^{-1}). Fossil fuels typically have high energy density compared to renewables.

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Efficiency is a measure of how much input energy is converted into useful output energy. In every energy transformation, some energy is 'dissipated' or 'wasted' as thermal energy (heat) to the surroundings.

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Sankey Diagrams are visual tools used to represent energy transformations. The width of the arrows is proportional to the amount of energy in Joules (JJ).

📐Formulae

Ek=12mv2E_k = \frac{1}{2}mv^2

Ep=mghE_p = mgh

W=F×dW = F \times d

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

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

Specific Energy=Energy released from fuelMass of fuel consumed\text{Specific Energy} = \frac{\text{Energy released from fuel}}{\text{Mass of fuel consumed}}

💡Examples

Problem 1:

A coal-fired power station consumes 2000 kg2000 \, kg of coal. The specific energy of coal is 30 MJ kg−130 \, MJ \, kg^{-1}. If the station produces 1.8×1010 J1.8 \times 10^{10} \, J of useful electrical energy, calculate its efficiency.

Solution:

First, calculate the total energy input from the coal: Total Input=mass×specific energy\text{Total Input} = \text{mass} \times \text{specific energy} Total Input=2000 kg×30×106 J kg−1=6.0×1010 J\text{Total Input} = 2000 \, kg \times 30 \times 10^6 \, J \, kg^{-1} = 6.0 \times 10^{10} \, J

Next, use the efficiency formula: Efficiency=1.8×1010 J6.0×1010 J×100%\text{Efficiency} = \frac{1.8 \times 10^{10} \, J}{6.0 \times 10^{10} \, J} \times 100\% Efficiency=0.3×100%=30%\text{Efficiency} = 0.3 \times 100\% = 30\%

Explanation:

The efficiency represents the percentage of chemical energy in the coal that was successfully converted into electrical energy. The remaining 70%70\% is lost as waste heat.

Problem 2:

A hydroelectric dam uses water falling from a height of 50 m50 \, m. If 2000 kg2000 \, kg of water flows every second, calculate the maximum theoretical power output.

Solution:

The power is the rate of change of Gravitational Potential Energy (EpE_p): ΔEp=mgh\Delta E_p = mgh ΔEp=2000 kg×9.8 m s−2×50 m\Delta E_p = 2000 \, kg \times 9.8 \, m \, s^{-2} \times 50 \, m ΔEp=980,000 J\Delta E_p = 980,000 \, J

Since this energy is released every second (t=1 st = 1 \, s): P=Et=980,000 J1 s=980,000 WP = \frac{E}{t} = \frac{980,000 \, J}{1 \, s} = 980,000 \, W P=980 kWP = 980 \, kW

Explanation:

The potential energy of the water at the top of the dam is converted into kinetic energy as it falls, which then spins turbines to generate electrical power.