Energy — Diagnostic Tests

Unit Tests

UT-1: Energy Stores and Transfers

Question: (a) Describe the energy transfers when a ball is thrown upwards and then caught at the same height. (b) A 0.5 kg ball is thrown vertically upwards at 10 m/s. Calculate the maximum height reached. (c) A 60 kg person runs up a flight of stairs 5 m high in 4 seconds. Calculate the power developed. (d) Explain why the actual height reached by the ball in part (b) would be less than calculated.

Solution:

(a) As the ball rises: kinetic energy $\to$ gravitational potential energy. At the top: maximum gravitational potential energy, zero kinetic energy (momentarily at rest). As it falls: gravitational potential energy $\to$ kinetic energy. When caught: kinetic energy $\to$ thermal energy (in the hand) and sound energy.

(b) $mgh = \frac{1}{2}mv^2$. $h = \frac{v^2}{2g} = \frac{100}{2 \times 9.8} = \frac{100}{19.6} = 5.10$ m.

(c) $P = \frac{E}{t} = \frac{mgh}{t} = \frac{60 \times 9.8 \times 5}{4} = \frac{2940}{4} = 735$ W.

(d) Air resistance does negative work on the ball, converting some kinetic energy to thermal energy. This means not all the initial kinetic energy is converted to gravitational potential energy, so the ball reaches a lower maximum height than predicted by the ideal calculation.

UT-2: Specific Heat Capacity

Question: (a) Define specific heat capacity and state its unit. (b) Calculate the energy needed to heat 2 kg of water from 20^\circ\text{C to 80^\circ\text{C. (Specific heat capacity of water = 4200 \text{ J/(kg ^\circ\text{C).) (c) A 0.5 kg metal block is heated from 25^\circ\text{C to 225^\circ\text{C using $45000$ J of energy. Calculate the specific heat capacity of the metal. (d) Explain why water has a much higher specific heat capacity than metals and why this is important for climate.

Solution:

(a) Specific heat capacity is the energy required to raise the temperature of 1 kg of a substance by 1^\circ\text{C. Unit: J/(kg $^\circ$C) or J/(kg K).

(b) $\Delta E = mc\Delta T = 2 \times 4200 \times (80 - 20) = 2 \times 4200 \times 60 = 504,000$ J $= 504$ kJ.

(c) $c = \frac{\Delta E}{m\Delta T} = \frac{45000}{0.5 \times (225 - 25)} = \frac{45000}{0.5 \times 200} = \frac{45000}{100} = 450$ J/(kg $^\circ$C).

(d) Water has strong hydrogen bonds between molecules, which must be partially broken to increase molecular motion (temperature). Metals have metallic bonding with free electrons that transfer energy . Water’s high SHC means it can absorb large amounts of energy with small temperature changes. This moderates coastal climates (oceans heat up slowly in summer, cool slowly in winter) and makes water effective as a coolant in engines and nuclear reactors.

UT-3: Efficiency

Question: (a) Define efficiency. (b) A power station burns $5 \times 10^8$ J of chemical energy per second and produces $1.5 \times 10^8$ W of electrical power. Calculate the efficiency. (c) Explain why it is impossible for a machine to have 100% efficiency. (d) A heat engine operates between a hot reservoir at 500^\circ\text{C and a cold reservoir at 25^\circ\text{C. Calculate the maximum theoretical efficiency.

Solution:

(a) Efficiency is the ratio of useful energy output to total energy input, expressed as a percentage. \text{Efficiency = \frac{\text{useful output}{\text{total input} \times 100\%.

(b) \text{Efficiency = \frac{1.5 \times 10^8}{5 \times 10^8} \times 100\% = 30\%.

(c) No machine can be 100% efficient because some energy is always dissipated as thermal energy due to friction between moving parts, air resistance, and sound. These waste energy transfers cannot be eliminated entirely. Additionally, the Second Law of Thermodynamics states that in any energy transfer, some energy is always dispersed in an unordered way (increasing entropy).

(d) Maximum theoretical efficiency $= 1 - \frac{T_c}{T_h}$ (using absolute temperatures in Kelvin). $T_c = 25 + 273 = 298$ K. $T_h = 500 + 273 = 773$ K. \text{Efficiency = 1 - \frac{298}{773} = 1 - 0.3855 = 0.6145 = 61.5\%.

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Integration Tests

IT-1: Energy Conservation in a Roller Coaster (with Forces)

Question: A roller coaster car of mass 800 kg starts from rest at point A, 30 m above the ground. At point B, the track is 10 m above the ground. (a) Calculate the speed of the car at point B, assuming no energy losses. (b) If the car’s actual speed at B is 18 m/s, calculate the energy lost to friction. (c) Calculate the average frictional force if the track from A to B is 120 m long. (d) Explain how the concept of gravitational potential energy and kinetic energy interchange relates to the conservation of energy principle.

Solution:

(a) By conservation of energy: $mgh_A = \frac{1}{2}mv_B^2 + mgh_B$. $800 \times 9.8 \times 30 = \frac{1}{2} \times 800 \times v_B^2 + 800 \times 9.8 \times 10$. $235,200 = 400v_B^2 + 78,400$. $400v_B^2 = 156,800$. $v_B^2 = 392$. $v_B = 19.8$ m/s.

(b) KE at actual speed: $\frac{1}{2} \times 800 \times 18^2 = 400 \times 324 = 129,600$ J. Expected KE: $156,800$ J. Energy lost $= 156,800 - 129,600 = 27,200$ J.

(c) Work done against friction $= F \times d = 27,200$. $F = 27,200 / 120 = 226.7$ N.

(d) As the coaster descends from A to B, gravitational potential energy (proportional to height) is converted to kinetic energy (proportional to speed squared). The total mechanical energy (GPE + KE) remains constant if no friction is present. This is the conservation of energy: energy cannot be created or destroyed, only transferred from one store to another. In reality, some mechanical energy is transferred to thermal energy by friction, but the total energy (including thermal) is always conserved.

IT-2: Power and Cost Calculations (with Electricity)

Question: A kettle rated at 2.2 kW is used to boil 1.5 kg of water from 18^\circ\text{C to 100^\circ\text{C. (a) Calculate the energy needed to heat the water. (b) Calculate the time taken if the kettle is 85% efficient. (c) If electricity costs 28p per kWh, calculate the cost of boiling the water. (d) A microwave rated at 800 W takes 3 minutes to heat the same water to 100^\circ\text{C. Calculate the efficiency of the microwave and compare the cost.

Solution:

(a) $\Delta E = mc\Delta T = 1.5 \times 4200 \times 82 = 516,600$ J $= 516.6$ kJ.

(b) Electrical energy input $= 516,600 / 0.85 = 607,765$ J $= 607.8$ kJ. Time $= \frac{607,765}{2200} = 276.3$ s $\approx 4.6$ minutes.

(c) Energy in kWh $= 607,765 / 3,600,000 = 0.1688$ kWh. Cost $= 0.1688 \times 28 = 4.73$ p.

(d) Microwave energy input $= 800 \times 180 = 144,000$ J. The useful energy required to heat 1.5 kg from 18^\circ\text{C to 100^\circ\text{C is 516.6 kJ, which exceeds the input of 144 kJ. This means the scenario as stated is physically impossible.

With a more realistic scenario: the microwave heats 0.3 kg from 18^\circ\text{C to 80^\circ\text{C in 3 minutes. $\Delta E = 0.3 \times 4200 \times 62 = 78,120$ J. Input $= 800 \times 180 = 144,000$ J. Efficiency $= 78,120/144,000 = 54.3\%$.

Cost of microwave $= 144,000 / 3,600,000 \times 28 = 0.04 \times 28 = 1.12$ p.

The kettle is more efficient (85% vs 54.3%) and costs 4.73p vs 1.12p, but the kettle heats more water to a higher temperature.

IT-3: Energy Resources and Environmental Impact (with Waves)

Question: (a) Compare solar and wind energy in terms of: energy source type, reliability, environmental impact, and typical output. (b) A solar panel has an area of 10 \text{ m^2 and an efficiency of 18%. If the solar irradiance is 800 \text{ W/m^2Calculate the power output. (c) A wind turbine has blades of length 40 m. If the wind speed is 12 m/s and the air density is 1.2 \text{ kg/m^3Calculate the maximum theoretical power (use $P = \frac{1}{2}\rho A v^3$ where $A = \pi r^2$). (d) Explain why no energy resource is completely free of environmental impact.

Solution:

(a)	Aspect	Solar	Wind
Type	Renewable (electromagnetic radiation)	Renewable (kinetic energy of air)
Reliability	Intermittent (day/night, weather dependent)	Intermittent (variable wind speeds)
Environmental impact	Land use, manufacturing (mining for silicon), disposal of panels	Visual impact, noise, bird/bat mortality, land use
Typical output	200—400 W per panel (residential)	2—5 MW per turbine (utility scale)

(b) Power = \text{irradiance \times \text{area \times \text{efficiency = 800 \times 10 \times 0.18 = 1440 W $= 1.44$ kW.

(c) A = \pi \times 40^2 = 5026.5 \text{ m^2. $P = \frac{1}{2} \times 1.2 \times 5026.5 \times 12^3 = 0.6 \times 5026.5 \times 1728 = 5,211,517$ W $\approx 5.2$ MW.

In practice, the Betz limit means a turbine can extract at most 59.3% of this: $5.2 \times 0.593 \approx 3.1$ MW.

(d) All energy resources have environmental impacts: solar panels require mining of rare earth metals and have disposal issues; wind turbines affect wildlife and landscapes; hydroelectric dams flood ecosystems and disrupt fish migration; nuclear produces radioactive waste; fossil fuels cause pollution and climate change. Even “clean” energy requires manufacturing, transportation, and infrastructure, all of which have carbon footprints and ecological impacts. The goal is to minimise rather than eliminate impact.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.