School playground microclimate survey: is it an urban heat island?
A Year 11 class investigates whether their school playground is an urban heat island. They record air temperature at six sites at 14:00 on three consecutive days in June using digital thermometers mounted on retort stands 1.2 m above ground. Site A is on dark tarmac in the middle of the playground; B is under a mature oak; C is beside a pond; D is on the grass sports field; E is next to a south-facing brick wall; F is in the car park beside the ventilation outlet of the kitchen. Results: A 31.4, 32.1, 31.8; B 26.2, 26.8, 26.5; C 27.0, 27.4, 27.1; D 28.1, 28.6, 28.3; E 33.0, 34.2, 33.5; F 35.1, 36.0, 35.6. The forecast for all three days was light winds and clear skies. One group's thermometer was dropped on day 2 and afterwards read 2.0 °C higher than a checked reference at every site where it was used; the group did not record which site readings used that thermometer.
1. Calculate the mean temperature for sites A and B. Explain why the difference between these means does not by itself prove an urban heat island effect.
Read the worked answer
Site A mean = (31.4 + 32.1 + 31.8) ÷ 3 = 95.3 ÷ 3 = 31.8 °C (1 d.p.). Site B mean = (26.2 + 26.8 + 26.5) ÷ 3 = 79.5 ÷ 3 = 26.5 °C. The difference is 5.3 °C.
This difference alone does not prove an urban heat island. A and B differ in shade, surface colour and moisture: B is under a mature oak, so it receives less direct solar radiation and may be cooler because of transpiration, not because the whole playground is urban. A is dark tarmac, which absorbs and re-radiates heat. The comparison is also only one pair of sites over three days, with no rural control site. An urban heat island is a difference between an urban area and its rural surroundings, so without a rural reference and repeated spatial sampling the 5.3 °C could simply be a local shade/surface effect.
2. The group concludes that site F is the hottest because of the urban heat island. Evaluate this conclusion.
Read the worked answer
Site F has the highest mean: (35.1 + 36.0 + 35.6) ÷ 3 = 106.7 ÷ 3 = 35.6 °C. That is 9.1 °C above site B. But the conclusion is weak because F is beside the kitchen ventilation outlet, a local heat source. It therefore measures exhaust heat, not the general urban surface effect. Site E is also beside a south-facing brick wall, which stores heat, and A is dark tarmac, so several sites have local confounders.
The dropped thermometer is a second problem: if it read 2.0 °C high and the group did not record where it was used, some site means may be inflated by an unknown amount. The conclusion cannot isolate the urban heat island from local heat sources and instrument error. A stronger design would move F away from the outlet, pair sites with similar surfaces, and use a calibrated thermometer at every site.
3. Recommend three changes to the fieldwork design to produce more reliable data.
Read the worked answer
- Calibrate and record instruments. Check every thermometer against a reference before and after fieldwork, number them, and record which thermometer is used at which site. This removes the ambiguity caused by the dropped thermometer.
- Control local heat sources and standardise exposure. Move site F away from the kitchen outlet and site E away from the wall, or add paired sites at the same surface type. Use a radiation shield or Stevenson screen at 1.2 m so direct sun does not bias readings.
- Increase spatial and temporal replication. Survey more sites, including a rural control outside the school grounds, and repeat at the same times on more days. Record wind speed, cloud cover and surface type at each site so the effect of weather and surface can be separated from an urban heat island signal.