MYP Physics · unit 18
Climate and energy systems
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
A planet's energy balance with an albedo feedback
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A planet receives stellar irradiance 1360 W m⁻² on a surface perpendicular to incoming light. Its albedo is initially 0.30. Assume a spherical planet and uniform effective emission σT⁴, where σ = 5.67×10⁻⁸ W m⁻² K⁻⁴. The supplied graph gives emitted flux for several effective temperatures. The actual surface is warmer than this simple emission temperature. Supplied data (Temperature / K; Emitted flux / W m⁻²): (240, 188.117), (255, 239.742), (270, 301.327), (288, 390.079). (a) Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four. [3 marks] (b) Albedo drops to 0.25. Find the new absorbed flux and temperature, and distinguish the greenhouse mechanism from the albedo change. [3 marks]
Blackbody emitted flux
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Emitted flux / W m⁻²
Horizontal axis (x): Temperature / K
Read the exact data as a table
| Temperature / K | Emitted flux / W m⁻² |
|---|---|
| 240 | 188.117 |
| 255 | 239.742 |
| 270 | 301.327 |
| 288 | 390.079 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Calculate globally averaged absorbed stellar flux and equilibrium effective temperature, explaining the factor of four.
Reveal the worked solution
- The planet intercepts light over a disc πR² but emits across a spherical area 4πR², so averaged incoming irradiance is S/4.
- Absorbed flux is (1−0.30)(1360)/4 = 238 W m⁻².
- Equilibrium requires σT⁴ = 238, giving T = (238/σ)¹/⁴≈254.6 K. This is the effective emission temperature, not automatically the surface temperature.
- With albedo 0.25, absorbed flux is 0.75(1360)/4 = 255 W m⁻², an increase of 17 W m⁻².
- New equilibrium temperature is (255/σ)¹/⁴≈259 K, about 4.4 K above the initial effective temperature.
- Albedo changes reflected incoming light. Greenhouse gases absorb and emit outgoing infrared radiation and alter energy transfer between surface and atmosphere. Melting reflective ice can lower albedo and reinforce warming; this simple model does not quantify the complete coupled climate response.
Answer (a) Absorbed flux = 238 W m⁻²; effective temperature≈254.6 K. (b) New flux = 255 W m⁻²; effective temperature≈259 K; lower albedo gives warming in this model.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A community energy plan that survives the demand peak
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A community uses 120 kWh of electricity per day, with the supplied six-hour average demand levels. A solar array generates 24 kWh per installed kW per week, and a wind installation generates 60 kWh per installed kW per week. A proposed plan uses 14 kW of solar plus 8.4 kW of wind. Treat the quoted outputs as weekly averages; neither source is dispatchable on demand. Supplied data (Period start / h; Average demand / kW): (0, 2), (6, 5), (12, 4), (18, 9). (a) Check daily demand and weekly energy balance. Explain why average energy balance is insufficient to establish reliability. [3 marks] (b) An evening backup battery must supply the full 9 kW for 6 h. Its usable depth of discharge is 80% and discharge efficiency is 90%. Find nominal energy capacity and state two further decision criteria. [3 marks]
Six-hour average demand
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Average demand / kW
Horizontal axis (x): Period start / h
Read the exact data as a table
| Period start / h | Average demand / kW |
|---|---|
| 0 | 2 |
| 6 | 5 |
| 12 | 4 |
| 18 | 9 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Check daily demand and weekly energy balance. Explain why average energy balance is insufficient to establish reliability.
Reveal the worked solution
- The four six-hour blocks consume 6(2+5+4+9) = 120 kWh per day, or 840 kWh per week.
- Proposed weekly generation is 14(24)+8.4(60) = 336+504 = 840 kWh, matching the average weekly energy.
- Matching totals does not align generation with the evening 9 kW demand or calm cloudy intervals. Storage, flexible demand or reliable backup is required, with losses and power limits included.
- Required delivered energy is 9(6) = 54 kWh. Usable delivered fraction is 0.80×0.90 = 0.72.
- Nominal energy capacity must be at least 54/0.72 = 75 kWh. This capacity calculation is separate from the required 9 kW output-power rating.
- Compare capital and lifecycle costs, resource variability, maintenance, local noise or land impacts, material supply and recycling. Stakeholder priorities and environmental effects should be stated explicitly rather than hidden in one energy total.
Answer (a) Daily demand = 120 kWh; generation = 840 kWh per week, equal to average demand, but peak reliability is unproven. (b) Nominal battery energy≥75 kWh; output power rating≥9 kW.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
A gas cycle with a sign-convention trap
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A fixed amount of gas follows the rectangular cycle A→B→C→D→A on the supplied pressure–volume data: A=(1 litre,100 kPa), B=(3 litre,100 kPa), C=(3 litre,300 kPa), D=(1 litre,300 kPa). Straight graph segments join the points. Use ΔU = Q−Wby, with work by the gas positive in expansion. For A→B the internal-energy increase is 300 J. Supplied data (Volume / litre; Pressure / kPa): (1, 100), (3, 100), (3, 300), (1, 300), (1, 100). (a) Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle. [3 marks] (b) Find net heat over the full cycle and assess whether this direction of operation is a heat engine. [3 marks]
Rectangular gas cycle
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Pressure / kPa
Horizontal axis (x): Volume / litre
Read the exact data as a table
| Volume / litre | Pressure / kPa |
|---|---|
| 1 | 100 |
| 3 | 100 |
| 3 | 300 |
| 1 | 300 |
| 1 | 100 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Find work and heat transferred on A→B, and determine net work by the gas over the complete cycle.
Reveal the worked solution
- At 100 kPa, expansion by 2 litres gives Wby = pΔV = 100000(0.002) = 200 J. Note that 1 kPa·litre = 1 J.
- The first law gives Q = ΔU+Wby = 300+200 = 500 J into the gas.
- On C→D, Wby = 300000(−0.002) = −600 J; vertical legs do no boundary work. Net work is 200−600 = −400 J, so 400 J is done on the gas.
- Internal energy is a state function, so returning to A gives ΔUcycle = 0.
- The first law then gives Qnet = Wby,net = −400 J: net energy leaves as heat while net work is supplied.
- A net-output heat engine would require the opposite orientation, but its efficiency cannot be found from the enclosed area alone; heat input on the relevant legs must also be known.
Answer (a) A→B: Wby = +200 J and Q = +500 J; net cycle Wby = −400 J. (b) Net Q = −400 J; this direction consumes net work rather than delivering heat-engine work.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
An engine's advertised efficiency exceeds its thermal limit
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A cyclic engine absorbs 1200 J per cycle from a 600 K reservoir and rejects heat to a 300 K reservoir. The manufacturer claims 800 J of work per cycle. The supplied graph shows the maximum reversible efficiency 1−Tc/Th as cold-reservoir temperature varies for Th = 600 K. Supplied model: For heat Q received reversibly by a reservoir at absolute temperature T, its entropy change is ΔS = Q/T, with Q negative for energy leaving that reservoir. Supplied data (Cold reservoir / K; Maximum efficiency): (150, 0.75), (300, 0.5), (450, 0.25), (600, 0). (a) Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle. [3 marks] (b) A real engine instead delivers 450 J per cycle. Find its efficiency and total reservoir entropy change; explain why waste heat cannot all be recycled into work in the same cycle. [3 marks]
Reversible efficiency limit
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Maximum efficiency
Horizontal axis (x): Cold reservoir / K
Read the exact data as a table
| Cold reservoir / K | Maximum efficiency |
|---|---|
| 150 | 0.75 |
| 300 | 0.5 |
| 450 | 0.25 |
| 600 | 0 |
These are the supplied values, not a worked answer. Lines join the supplied values; they are not a fitted model.
Show a hint
Begin with part (a): Check the claim using both energy conservation and the second law. Find maximum permitted work and minimum rejected heat per cycle.
Reveal the worked solution
- Energy conservation alone allows Qout = 1200−800 = 400 J, so a first-law calculation does not expose the problem.
- The reversible limit is 1−300/600 = 0.50. Maximum work is 0.50(1200) = 600 J, and at least 600 J must be rejected.
- At the claimed output the reservoir entropy change would be −1200/600+400/300 = −2/3 J K⁻¹ per cycle, violating the nondecrease of total entropy.
- The actual rejected heat is 1200−450 = 750 J. Efficiency is 450/1200 = 0.375.
- Reservoir entropy change is −1200/600+750/300 = −2+2.5 = +0.50 J K⁻¹ per cycle, consistent with irreversibility.
- The working substance returns to its initial state each cycle. Converting all heat from a single reservoir into cyclic work would violate the second law; reducing losses cannot remove the need for heat rejection at these temperatures.
Answer (a) Claimed efficiency = 66.7% exceeds the 50% limit; Wmax = 600 J and Qrejected,min = 600 J. (b) Efficiency = 37.5%; reservoir entropy increase = 0.50 J K⁻¹ per cycle.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.