S5 Geography Paper: Weather and Climate of Hong Kong (100 marks)

2026-11-12

Geography – Weather and Climate

S5 Geography: 6 questions, 24 parts, 100 marks, 75 minutes. Pressure gradient and wind, humidity, tropical cyclones, cold fronts, rainfall data and urban heat island.

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Question 1 Pressure Gradient and Wind (16 marks)

Study the data below and answer all parts.

#	Question	Answer and marking notes
1	On a certain day the Hong Kong station records 1012.4 hPa while a station 95 km to its northwest records 1004.8 hPa. Calculate the pressure gradient in hPa per km (2 d.p.) and convert it to a value per 100 km.	Pressure difference = 1012.4 – 1004.8 = 7.6 hPa; gradient = 7.6 / 95 = 0.08 hPa/km; per 100 km = 8.0 hPa. (4 marks: difference 1, division 1, 0.08 1, 8.0 1)
2	If the two stations were only 40 km apart with the same 7.6 hPa difference, what would the gradient be and how would the wind between them change?	7.6 / 40 = 0.19 hPa/km, about 2.4 times the original; the pressure gradient force is stronger so wind speed increases. (4 marks)
3	Explain how the pressure gradient force, the Coriolis force and friction together determine surface wind direction and speed in the northern hemisphere.	The pressure gradient force acts from high to low pressure and sets wind speed; the Coriolis force deflects wind to the right; friction reduces speed and turns wind across the isobars at about 20 to 30 degrees, giving a three-force balance. (4 marks)
4	Compare the wind speeds in parts (a) and (b) using the gradient ratio, and explain why the southwest monsoon prevails in Hong Kong in summer.	The ratio is about 0.19 / 0.08 = 2.4 times; in summer the heated Asian landmass forms a low pressure centre, so the gradient force points inland from the ocean and, with the Coriolis deflection, Hong Kong receives moist southwest winds. (4 marks)

Question 2 Temperature, Humidity and Rising Air (20 marks)

Answer each part independently and show your working.

#	Question	Answer and marking notes
5	At the airport the air temperature is 32.5 C with a vapour pressure of 34.2 hPa; the saturation vapour pressure at that temperature is 47.5 hPa. Calculate the relative humidity (1 d.p.).	Relative humidity = 34.2 / 47.5 x 100% = 72.0%. (4 marks)
6	The dew point at the station is 26.5 C. Calculate the dew point depression and state the temperature at which the air becomes saturated.	Dew point depression = 32.5 – 26.5 = 6.0 C; the air becomes saturated and condensation begins at 26.5 C. (4 marks)
7	Unsaturated air rises at the dry adiabatic lapse rate of 10 C per km. How much does the temperature fall after rising 500 m, and what is the temperature at that height if the surface temperature is 29.0 C?	Fall = 10 x 0.5 = 5.0 C; temperature = 29.0 – 5.0 = 24.0 C. (4 marks)
8	The Tai Mo Shan summit is 957 m high; the environmental lapse rate is 6.5 C per km and the surface temperature is 29.0 C. Calculate the summit temperature (1 d.p.) and explain why cloud forms more readily there.	Fall = 6.5 x 0.957 = 6.2 C; summit temperature = 29.0 – 6.2 = 22.8 C; the cooler air reaches saturation more easily, so cloud forms. (4 marks)
9	If the temperature rises to 35.0 C (saturation vapour pressure 56.2 hPa) while vapour pressure stays at 34.2 hPa, what is the new relative humidity and why does it fall?	Relative humidity = 34.2 / 56.2 x 100% = 60.9%; a higher temperature raises the saturation vapour pressure, so the same vapour content gives a smaller ratio. (4 marks)

Question 3 Tropical Cyclones (18 marks)

The sea level pressure, wind and rainfall data of a tropical cyclone are given below.

#	Question	Answer and marking notes
10	The cyclone centre has a pressure of 950 hPa and the pressure is 1010 hPa at 200 km from the centre. Calculate the mean pressure gradient (hPa/km) and explain why winds are strongest near the centre.	Gradient = (1010 – 950) / 200 = 0.30 hPa/km; isobars are closest near the centre, so the pressure gradient force and wind speed are greatest there. (5 marks)
11	28 mm of rain is recorded in 10 minutes. Calculate the rainfall intensity in mm per hour and the total rainfall if the intensity lasts 3 hours.	Intensity = 28 x 6 = 168 mm/h; total = 168 x 3 = 504 mm. (5 marks)
12	The maximum wind speed near the centre increases from 120 km/h to 165 km/h. Calculate the percentage increase and the change in wind energy, which is proportional to the square of wind speed.	Increase = (165 – 120) / 120 x 100% = 37.5%; energy ratio = (165 / 120)^2 = 1.89 times. (4 marks)
13	The astronomical tide is 2.3 m, the storm surge adds 1.4 m and the sea wall is 3.5 m high. Calculate the peak water level and state whether overtopping occurs.	Peak level = 2.3 + 1.4 = 3.7 m, which is 0.2 m above the wall, so overtopping and coastal flooding occur. (4 marks)

Question 4 Passage of a Cold Front (16 marks)

A cold front moves southeast; the observations before and after its passage are shown below.

#	Question	Answer and marking notes
14	The cold front is 180 km northwest of Hong Kong and moves southeast at 30 km/h. How many hours will it take to reach Hong Kong?	Time = 180 / 30 = 6 hours. (4 marks)
15	The temperature is 27.5 C before the front passes and 19.0 C after. Calculate the temperature drop.	Drop = 27.5 – 19.0 = 8.5 C. (4 marks)
16	45 mm of rain falls in the 3 hours of frontal passage. Calculate the mean intensity; if that intensity lasts the first 6 hours and the remaining 18 hours give 2 mm each, find the daily total.	Intensity = 45 / 3 = 15 mm/h; daily total = 15 x 6 + 2 x 18 = 90 + 36 = 126 mm. (4 marks)
17	Relative humidity is 88% before the front and 62% after. Calculate the fall in percentage points and explain why the weather becomes fine after a cold front passes.	Fall = 88 – 62 = 26 percentage points; the cold air mass is dry with subsiding air, so cloud disperses and visibility improves. (4 marks)

Question 5 Rainfall Data Analysis (16 marks)

Monthly rainfall (mm) at a station for six months is 18, 32, 45, 210, 385 and 402.

#	Question	Answer and marking notes
18	Calculate the total rainfall for the six months and the mean monthly rainfall (1 d.p.).	Total = 18 + 32 + 45 + 210 + 385 + 402 = 1,092 mm; mean = 1,092 / 6 = 182.0 mm. (4 marks)
19	Find the median of the six monthly values and explain why the median represents dry season rainfall better than the mean.	Sorted: 18, 32, 45, 210, 385, 402; median = (45 + 210) / 2 = 127.5 mm; the median is not inflated by the extreme wet months. (4 marks)
20	Calculate the wettest month as a percentage of the six-month total (1 d.p.).	402 / 1,092 x 100% = 36.8%. (4 marks)
21	Calculate the share of the total taken by the three wettest months (210, 385, 402 mm) and state one impact of a concentrated wet season on urban drainage.	997 / 1,092 x 100% = 91.3%; a concentrated wet season produces a sharp runoff peak that can exceed drainage capacity and cause flooding. (4 marks)

Question 6 Urban Heat Island and Climate Change (14 marks)

The data show urban and rural temperatures and long term annual mean temperatures for Hong Kong.

#	Question	Answer and marking notes
22	The urban temperature is 33.4 C and the nearby rural temperature is 29.1 C. Calculate the heat island intensity and suggest two planning measures to reduce it.	Intensity = 33.4 – 29.1 = 4.3 C; measures include more urban greenery and green roofs, and keeping air ventilation corridors open. (5 marks)
23	The mean temperature was 23.1 C in 1991 to 2000 and 24.0 C in 2011 to 2020. Calculate the total rise and the mean rise per decade, and explain the roles of urbanisation and global warming.	Total rise = 24.0 – 23.1 = 0.9 C over two decades, that is 0.45 C per decade; urbanisation adds heat storage and reduces evapotranspiration while global warming raises the regional baseline, and the two combine. (5 marks)
24	If the trend of 0.45 C per decade continues, estimate the temperature in 2030 relative to 24.0 C in 2020 and give one limitation of the estimate.	2030 temperature = 24.0 + 0.45 = 24.45 C, about 24.5 C; limitations include ignoring emission cuts, natural variability and changes in the pace of urbanisation. (4 marks)

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