S5 Geography Practice Paper: Weather, Climate And Urban Climate (12 Questions, 100 Marks, 75 Minutes, With Answers)

S5 Geography Practice Paper: Weather, Climate and Urban Climate (12 Questions, 100 Marks, 75 Minutes, with Answers)

2026-11-06

HKDSE Geography: Weather, Climate and Urban Climate

S5 geography practice paper, 12 questions, 100 marks, 75 minutes: lapse rates, relative humidity and condensation, rainfall distribution, tropical cyclones, urban heat island, radiation bal

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Section A: Temperature, Humidity and Precipitation (24 marks)

Each question carries 6 marks; show your working or reasoning.

#	Question	Answer and marking notes
1	On a certain day the temperature at the foot of a mountain (0 m) is 28.0 C and a weather station on the summit is at 1,200 m. (a) If the rising air is unsaturated and the dry adiabatic lapse rate is 6.5 C per 1,000 m, calculate the summit temperature. (b) If the air is saturated and the saturated adiabatic lapse rate is 5.0 C per 1,000 m, calculate the summit temperature. (c) Explain why moist air cools more slowly as it rises.	(a) Summit temperature = 28.0 – 6.5 x 1.2 = 20.2 C. (2 marks) (b) Summit temperature = 28.0 – 5.0 x 1.2 = 22.0 C. (2 marks) (c) Moist air cools to its dew point as it rises, and condensation releases latent heat that offsets part of the cooling, so it cools by less per 1,000 m than dry air. (2 marks)
2	At 25 C the saturation vapour pressure is 31.7 hPa and the actual vapour pressure of the air is 19.0 hPa. (a) Calculate the relative humidity to one decimal place. (b) If the air cools to 17 C, where the saturation vapour pressure is 19.4 hPa, decide whether condensation will occur and explain. (c) Explain why walls and floors become wet during the humid spring spell in Hong Kong.	(a) Relative humidity = 19.0 / 31.7 x 100% = 59.9%. (2 marks) (b) No condensation: the actual vapour pressure of 19.0 hPa is below the saturation vapour pressure of 19.4 hPa at 17 C, so the air is still unsaturated; it must cool below its dew point before condensation occurs. (2 marks) (c) Warm moist air from the south meets walls and floors that are still cold, so the air next to those surfaces cools below its dew point and water vapour condenses on them. (2 marks)
3	The monthly rainfall totals of a station, in millimetres, are 15, 28, 45, 120, 290, 420, 380, 340, 210, 60, 25 and 18. (a) Calculate the annual total. (b) Calculate the percentage of the annual total that falls in the wettest month, to one decimal place. (c) State the climate type suggested by this seasonal distribution and give one agricultural risk it creates.	(a) Annual total = 1,951 mm. (2 marks) (b) 420 / 1,951 x 100% = 21.5%. (2 marks) (c) Rainfall concentrated in summer with a dry winter indicates a monsoon or tropical wet-dry climate; a late or very concentrated wet season can cause landslides on slopes and crop failure. (2 marks)
4	A tropical cyclone has a central pressure of 950 hPa, an environmental pressure of 1010 hPa and a ten-minute mean wind speed near the centre of 185 km/h. (a) Calculate the pressure deficit between the centre and the environment. (b) Convert the wind speed to m/s, to one decimal place. (c) Explain why a tropical cyclone weakens rapidly after landfall.	(a) Pressure deficit = 1010 – 950 = 60 hPa. (2 marks) (b) 185 / 3.6 = 51.4 m/s. (2 marks) (c) After landfall the storm loses the supply of water vapour and latent heat from the warm sea, while surface friction increases and terrain disrupts the circulation, so it weakens and fills rapidly. (2 marks)

Section B: Urban Climate, Radiation and Wind (36 marks)

Each question carries 9 marks; note the marks for each part.

#	Question	Answer and marking notes
5	An urban station records 32.4 C while a nearby rural station records 28.9 C. (a) Calculate the urban heat island intensity. (b) Research shows that each increase of 10 percentage points in urban greenery lowers summer daytime temperature by 0.6 C; if greenery rises from 12% to 25%, calculate the expected cooling. (c) State three causes of the urban heat island and suggest two mitigation measures.	(a) Heat island intensity = 32.4 – 28.9 = 3.5 C. (3 marks) (b) Greenery rises by 13 percentage points, so cooling = 13 / 10 x 0.6 = 0.78 C. (3 marks) (c) Causes: dense buildings and asphalt absorb heat and reduce evapotranspiration; anthropogenic heat from air conditioning and vehicles; tall buildings block ventilation and slow the loss of long-wave radiation at night. Measures: add green roofs, street trees and water bodies; keep ventilation corridors open and use pale high-albedo surfaces. (3 marks)
6	An air parcel at the surface has a temperature of 26 C and a dew point of 18 C, and rises at the dry adiabatic lapse rate of 6.5 C per 1,000 m while unsaturated. (a) Calculate how far the parcel must rise to cool to its dew point, that is the condensation level. (b) State the relationship between the condensation level and cloud base. (c) Explain why cloud bases are usually higher on afternoons with strong summer convection.	(a) Condensation level = (26 – 18) / 6.5 x 1000 = 1,231 m. (3 marks) (b) Without an inversion or dry layer the cloud base is at about the condensation level, because water vapour begins to condense into droplets at that height. (3 marks) (c) Strong afternoon heating gives a large gap between the starting temperature and the dew point, so the parcel must rise further before it cools to the dew point; cloud base is therefore higher and convective cloud grows thicker. (3 marks)
7	Mean incoming solar radiation at the top of the atmosphere is 340 W/m2 and the surface albedo is 30%. (a) Calculate the radiation reflected back to space. (b) Calculate the radiation absorbed by the Earth system. (c) Explain how greenhouse gases make the mean surface temperature higher than it would be without an atmosphere.	(a) Reflected = 340 x 30% = 102 W/m2. (3 marks) (b) Absorbed = 340 – 102 = 238 W/m2. (3 marks) (c) Greenhouse gases absorb the long-wave radiation emitted by the surface and re-emit it in all directions, some of it back towards the ground; net loss from the surface is reduced and the equilibrium temperature rises. Without an atmosphere the mean surface temperature would be much lower. (3 marks)
8	Two isobars differ by 4 hPa and are 200 km apart. (a) Calculate the pressure gradient in hPa per 100 km. (b) If the pressure difference rises to 6 hPa over the same distance, calculate the new gradient and the factor by which the wind speed would increase. (c) Explain how the Coriolis force and surface friction affect the direction and speed of surface wind.	(a) Pressure gradient = 4 / 200 x 100 = 2.0 hPa per 100 km. (3 marks) (b) New gradient = 6 / 200 x 100 = 3.0 hPa per 100 km, about 1.5 times the original, so with other factors unchanged the wind speed would also rise about 1.5 times. (3 marks) (c) The Coriolis force turns the wind away from the pressure gradient direction, to the right in the northern hemisphere, while surface friction slows the wind and turns it further so that it crosses the isobars towards low pressure; surface wind is therefore not perpendicular to the isobars. (3 marks)

Section C: Storms, Monsoon, Air Quality and Climate Change (40 marks)

Each question carries 10 marks; give full explanations or evaluations.

#	Question	Answer and marking notes
9	A catchment of 2.5 square kilometres receives 80 mm of rain in one hour and has a runoff coefficient of 0.6. (a) Calculate the volume of runoff generated in that hour, in cubic metres. (b) The Black Rainstorm signal is issued at 70 mm in one hour; decide which warning level this rainfall reaches. (c) Explain how urbanisation changes the runoff volume and the timing of the flood peak for the same storm.	(a) Runoff volume = 0.080 x 2,500,000 x 0.6 = 120,000 cubic metres. (4 marks) (b) 80 mm in one hour exceeds the 70 mm Black Rainstorm threshold, so the rainfall reaches the Black Rainstorm level. (3 marks) (c) Concrete and paved surfaces reduce infiltration and raise the runoff coefficient, so the runoff volume rises; drains and channels concentrate the water, so the flood peak arrives sooner and is higher. (3 marks)
10	Hong Kong experiences prevailing northerly winds in winter and south-westerly winds in summer. (a) Name this seasonal reversal of wind direction. (b) Explain its cause using the thermal contrast between land and sea and the seasonal shift of pressure belts. (c) Explain how the winter and summer monsoons affect the temperature and rainfall of Hong Kong differently.	(a) This is the monsoon, a seasonal reversal of wind direction. (3 marks) (b) In winter the land cools faster than the sea, so the interior forms a high and the sea a relative low, and wind blows from land to sea; in summer the land heats faster and forms a low, so wind blows from sea to land. The seasonal shift of the planetary wind and pressure belts towards and away from the hemisphere in summer reinforces these opposing flows. (4 marks) (c) The winter monsoon comes from cooler continental interiors, so Hong Kong is cool and dry with little rain; the summer monsoon comes from the ocean, warm and moist, and brings convection and monsoon rain that supplies most of the annual rainfall. (3 marks)
11	On one day the mixing depth is 800 m with a mean wind speed of 3 m/s; on another day the mixing depth is 400 m with a mean wind speed of 1.5 m/s. (a) Calculate the ventilation index in m2/s for both days. (b) Decide on which day pollutants accumulate more readily and explain. (c) Suggest two measures to reduce roadside pollution on days with poor ventilation.	(a) First day = 800 x 3 = 2,400 m2/s; second day = 400 x 1.5 = 600.0 m2/s. (4 marks) (b) Pollutants accumulate more on the second day: the ventilation index is only a quarter of that on the first day, so the shallow mixing layer and light wind dilute and disperse pollution inefficiently near the ground. (3 marks) (c) Restrict private cars in busy districts and raise the share of public transport and electric vehicles; increase road washing and site watering, and limit wall-like high-rise development along ventilation corridors. (3 marks)
12	The thirty-year mean temperature at a station for 1991 to 2020 is 23.3 C and the mean for the last ten years is 24.0 C. (a) Calculate the difference between the two periods. (b) If the rise is concentrated in the last twenty years, calculate the mean rise per decade. (c) State two limitations of inferring global warming from the temperature record of a single station.	(a) Difference = 24.0 – 23.3 = 0.7 C. (4 marks) (b) 0.7 / 2 = 0.35 C per decade. (3 marks) (c) Limitations: a single station is affected by relocation, instrument changes and the urban heat island, so it may not represent a regional trend; the thirty-year and ten-year means cover different periods and are influenced by year-to-year variability such as El Nino, so the conclusion must be checked against multi-station and global gridded data. (3 marks)

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