S5 Geography Paper: Climate Systems, Extreme Weather And Natural Hazards (100 Marks)

S5 Geography Paper: Climate Systems, Extreme Weather and Natural Hazards (100 marks)

2026-11-11

S5 Geography – climate and hazards

S5 Geography paper: pressure gradient and wind, lapse rates, humidity and dew point, greenhouse forcing, floods and seismic waves, with answers.

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Paper details　S5 Geography practice paper with six structured questions on pressure gradient and geostrophic wind, environmental and adiabatic lapse rates in mountain climates, relative humidity, dew point and condensation, greenhouse gas radiative forcing, flood return period and peak discharge, and seismic wave travel times with moment magnitude, with full answers and marking notes.

Part 1: Pressure gradient and geostrophic wind (16 marks)

On a certain day Station A records 1016 hPa and Station B records 1012 hPa; the two stations are 400 km apart, both at latitude 22.5 degrees north. Air density is 1.2 kg per cubic metre and the angular velocity of the Earth is 7.292 x 10^-5 radian per second. Show your working for every part.

#	Question	Answer and marking notes
1	Find the pressure gradient between the two stations in hPa per 100 km and in Pa per metre.	Answer: Pressure difference = 1016 – 1012 = 4 hPa. Gradient = 4 / 400 x 100 = 1.0 hPa per 100 km, that is 400 Pa / 400 000 m = 0.001 Pa per metre. (Marking: 1 mark for the difference, 2 marks for hPa per 100 km, 2 marks for Pa per metre.)
2	Find the Coriolis parameter f and the geostrophic wind speed in m/s to one decimal place.	Answer: f = 2 x omega x sin(latitude) = 2 x 7.292 x 10^-5 x sin 22.5 = 5.58 x 10^-5 per second. Geostrophic wind V = (dp/dn) / (density x f) = 0.001 / (1.2 x 5.58 x 10^-5) = 14.9 m/s. (Marking: 2 marks for f, 1 mark for the formula, 2 marks for substitution and answer.)
3	Convert that geostrophic wind speed into km/h and state one reason why the actual surface wind is usually weaker.	Answer: 14.9 x 3.6 = 53.8 km/h. Surface friction reduces the effect of the pressure gradient force, so the wind is slower and crosses the isobars at an angle. (Marking: 2 marks for the conversion, 1 mark for the reason.)
4	Explain how the pressure gradient force and the Coriolis force together determine the direction of the geostrophic wind.	Answer: The pressure gradient force acts from high to low pressure, while the Coriolis force in the northern hemisphere acts at right angles to the wind, to its right. When the two are equal and opposite the wind blows along the isobars with low pressure on the left and high pressure on the right, which is the geostrophic wind. (Marking: 1 mark for each force direction, 1 mark for the balance condition.)

Part 2: Lapse rates and mountain climate (16 marks)

A tropical coastal city has a sea level temperature of 28.0 C and a dew point of 18.0 C, with an environmental lapse rate of 6.5 C per km. A moist air parcel is lifted by the terrain from sea level; the dry adiabatic lapse rate is 10 C per km and the saturated adiabatic lapse rate is 6 C per km. Take the lifting condensation level as 125 m for each degree of dew point depression.

#	Question	Answer and marking notes
5	Find the environmental temperature at 2 500 m above sea level.	Answer: Temperature = 28.0 – 6.5 x 2.5 = 28.0 – 16.25 = 11.8 C. (Marking: 2 marks for the temperature drop, 2 marks for the answer.)
6	Find the height of the 0 C isotherm and explain its relationship to the snow line.	Answer: Height = 28.0 / 6.5 = 4.3077 km, about 4308 m. With no other factors the snow line lies close to the 0 C isotherm, so the snow line here is near 4 300 m. (Marking: 3 marks for the calculation, 1 mark for the relationship.)
7	Find the lifting condensation level of the parcel and the temperature there.	Answer: Lifting condensation level = (28.0 – 18.0) x 125 = 1250 m. Temperature there = 28.0 – 10 x 1.25 = 15.5 C. (Marking: 2 marks for the height, 2 marks for the temperature.)
8	Find the parcel temperature at the summit (2 500 m) and compare it with the environmental temperature at the same height.	Answer: Summit temperature = 15.5 – 6 x 1.25 = 8.0 C, which is 3.75 C lower than the environmental temperature of 11.75 C, so the parcel is colder than its surroundings. (Marking: 3 marks for the calculation, 1 mark for the comparison.)

Part 3: Relative humidity, dew point and condensation (18 marks)

Indoor air is at 30.0 C with a saturation vapour pressure of 42.4 hPa and an actual vapour pressure of 25.4 hPa. Use absolute humidity (g per cubic metre) = 217 x e / (T + 273.15), where e is in hPa and T in C. The saturation vapour pressure at 20.0 C is 23.4 hPa.

#	Question	Answer and marking notes
9	Find the relative humidity.	Answer: Relative humidity = 25.4 / 42.4 x 100 = 59.9 per cent. The vapour pressure is below saturation, so the air is unsaturated. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the answer.)
10	Estimate the dew point using 1 C of dew point depression for each 5 per cent of relative humidity deficit.	Answer: Humidity deficit = 100 – 59.9 = 40.1 per cent, so the dew point = 30.0 – 40.1 / 5 = 30.0 – 8.02 = 21.98 C, about 22.0 C. (Marking: 2 marks for the deficit, 2 marks for the answer.)
11	Find the water content of each cubic metre of air (absolute humidity).	Answer: Absolute humidity = 217 x 25.4 / 303.15 = 18.18 g per cubic metre. (Marking: 2 marks for the formula, 2 marks for the answer.)
12	If the air is cooled at constant pressure to 20.0 C, decide whether condensation occurs and find the mass of water condensed per cubic metre.	Answer: Saturated water content at 20.0 C = 217 x 23.4 / 293.15 = 17.32 g per cubic metre, and the actual 18.18 g per cubic metre exceeds it, so condensation occurs. Condensed mass = 18.18 – 17.32 = 0.86 g per cubic metre, forming fog or dew. (Marking: 2 marks for the saturation value, 1 mark for the decision, 2 marks for the condensed mass.)

Part 4: Greenhouse gases and radiative forcing (16 marks)

Atmospheric carbon dioxide rose from 280 ppm in 1750 to 420 ppm in 2020. Radiative forcing can be estimated as dF = 5.35 x ln(C/C0) in W per square metre. Take each 1 ppm of carbon dioxide as equivalent to 7.81 Gt of carbon dioxide emissions, and recent growth as about 2.5 ppm per year.

#	Question	Answer and marking notes
13	Find the increase in carbon dioxide concentration from 1750 to 2020 and the percentage increase.	Answer: Increase = 420 – 280 = 140 ppm, and the percentage increase = 140 / 280 x 100 = 50 per cent. (Marking: 2 marks for the increase, 2 marks for the percentage.)
14	Find the radiative forcing due to carbon dioxide between 1750 and 2020.	Answer: dF = 5.35 x ln(420 / 280) = 5.35 x ln 1.5 = 5.35 x 0.4055 = 2.17 W per square metre; a positive value means energy is retained in the climate system. (Marking: 2 marks for the formula, 2 marks for substitution and answer.)
15	If concentration rises 2.5 ppm each year, find the number of years needed to reach 560 ppm from 420 ppm.	Answer: Years needed = (560 – 420) / 2.5 = 140 / 2.5 = 56 years. (Marking: 2 marks for the concentration difference, 2 marks for the answer.)
16	Find the cumulative carbon dioxide emissions corresponding to that concentration increase between 1750 and 2020.	Answer: Cumulative emissions = 140 x 7.81 = 1093 Gt of carbon dioxide. Only part of total emissions has stayed in the air, the rest being taken up by the ocean and the land. (Marking: 3 marks for the calculation, 1 mark for the explanation.)

Part 5: Flood return period and peak discharge (18 marks)

A low-lying village sits at the outlet of a catchment with a design flood return period of 50 years. The catchment is 2.5 square kilometres, storm intensity is 120 mm per hour and the rural runoff coefficient is 0.65. Use the rational formula Q = 0.278 x C x i x A, with Q in cubic metres per second, i in mm per hour and A in square kilometres. The downstream channel is 30 m wide and carries water 2.0 m deep at a mean velocity of 1.8 m per second.

#	Question	Answer and marking notes
17	Find the annual exceedance probability of the design flood.	Answer: Annual exceedance probability = 1 / 50 = 0.02, that is 2 per cent in any year. (Marking: 2 marks for the formula, 2 marks for the percentage.)
18	Find the probability of at least one such flood in the next 30 years.	Answer: P = 1 – (1 – 0.02)^30 = 1 – 0.98^30 = 1 – 0.5455 = 0.45, about 45.45 per cent, close to one half. (Marking: 2 marks for the formula, 2 marks for the calculation, 1 mark for the interpretation.)
19	Find the design peak discharge before urbanisation.	Answer: Q = 0.278 x 0.65 x 120 x 2.5 = 54.21 cubic metres per second. (Marking: 2 marks for the formula, 2 marks for substitution, 1 mark for the answer.)
20	Find the channel capacity at that depth and decide whether it can cope after urbanisation raises the runoff coefficient to 0.85.	Answer: Capacity = 30 x 2.0 x 1.8 = 108 cubic metres per second. After urbanisation Q = 0.278 x 0.85 x 120 x 2.5 = 70.89 cubic metres per second, still below 108, so the channel copes, but the safety margin falls from 2.0 times to 1.5 times. (Marking: 2 marks for the capacity, 1 mark for the urban discharge, 1 mark for the decision.)

Part 6: Seismic wave travel time and moment magnitude (16 marks)

A seismic station records a 40 s difference between the arrivals of the P and S waves; crustal velocities there are 6.5 km per second for P waves and 3.7 km per second for S waves. The rupture plane is 20 km long and 15 km wide with a mean slip of 2.5 m, and the shear modulus is 3.0 x 10^10 Pa. Use Mw = (2/3) x (log10 M0 – 9.1) with M0 in N m.

#	Question	Answer and marking notes
21	Find the epicentral distance.	Answer: Let the distance be d. Then d / 3.7 – d / 6.5 = 40, so d x (0.27027 – 0.15385) = 40 and d = 343.6 km. (Marking: 2 marks for the equation, 2 marks for the calculation.)
22	Find the travel times of the P and S waves from the focus to the station.	Answer: S wave travel time = 343.6 / 3.7 = 92.9 s and P wave travel time = 343.6 / 6.5 = 52.9 s; the difference of 40 s matches the record. (Marking: 2 marks for each travel time.)
23	Find the area of the rupture plane and the seismic moment M0.	Answer: Area = 20 x 15 = 300 square kilometres = 3.0 x 10^8 square metres. M0 = 3.0 x 10^10 x 3.0 x 10^8 x 2.5 = 2.25e+19 N m. (Marking: 2 marks for the area conversion, 2 marks for M0.)
24	Find the moment magnitude Mw and explain why large earthquakes are usually described by Mw rather than the Richter scale.	Answer: Mw = (2/3) x (log10 2.25e+19 – 9.1) = (2/3) x (19.35 – 9.1) = (2/3) x 10.25 = 6.83. The Richter scale saturates above about magnitude 7 and cannot separate larger events, whereas Mw measures the energy released directly, so it is used for large earthquakes. (Marking: 2 marks for Mw, 2 marks for the explanation.)

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