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City in Georgia, United States From Wikipedia, the free encyclopedia
Summertown is a city in Emanuel County, Georgia, United States. The population was 121 in 2020.
Summertown, Georgia | |
---|---|
![]() Location in Emanuel County and the state of Georgia | |
Coordinates: 32°44′48″N 82°16′34″W | |
Country | United States |
State | Georgia |
County | Emanuel |
Area | |
• Total | 0.79 sq mi (2.05 km2) |
• Land | 0.79 sq mi (2.04 km2) |
• Water | 0.00 sq mi (0.01 km2) |
Elevation | 253 ft (77 m) |
Population (2020) | |
• Total | 121 |
• Density | 153.94/sq mi (59.40/km2) |
Time zone | UTC-5 (Eastern (EST)) |
• Summer (DST) | UTC-4 (EDT) |
ZIP code | 30401 |
Area code | 478 |
FIPS code | 13-74348[2] |
GNIS feature ID | 0323743[3] |
Summertown was originally built up as a summer retreat, hence the name.[4] The Georgia General Assembly incorporated Summertown as a town in 1906.[5]
Summertown is located in northern Emanuel County at 32°44′48″N 82°16′34″W (32.746532, -82.276182).[6] Georgia State Route 56 passes through the east side of the city limits, leading north 5 miles (8 km) to Midville and south 11 miles (18 km) to Swainsboro, the county seat.
According to the United States Census Bureau, Summertown has a total area of 0.77 square miles (2.0 km2), of which 0.004 square miles (0.01 km2), or 0.54%, is water.[7]
As of the census[2] of 2000, there were 140 people, 48 households, and 34 families residing in the town. By 2020, its population declined to 121.
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