Port Jefferson

Village in Shelby County, Ohio

View of Port Jefferson, Ohio
Photo: Nyttend · Public domain

Port Jefferson is a village in Shelby County, Ohio, with 308 residents (2020 Census). It sits at 971 ft (296 m) above sea level. The nearest town is Sidney, 4.9 miles SW.

About Port Jefferson

Port Jefferson is a village in Salem Township, Shelby County, Ohio, United States. The population was 308 at the 2020 census.

Text: Wikipedia · CC BY-SA 4.0

Port Jefferson weather: 7-day forecast

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Source: National Weather Service (NOAA).

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Living in Port Jefferson

Median household income
$57,083
Median home value
$101,000
Median rent
$850/mo
Median age
51.1
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Map of Port Jefferson

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Port Jefferson Street View

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Port Jefferson, OH population

Port Jefferson had 308 residents in the 2020 Census.

2020 Census308
Density2,020 per sq mi (780 per km²)

Source: U.S. Census Bureau, Decennial Census.

Port Jefferson at a glance

StateOhio (OH)
CountyShelby County
TypeVillage
Population308 (2020)
Land area0.15 sq mi (0.4 km²)
Water area0.00 sq mi (0.00 km²)
Elevation971 ft (296 m)
Coordinates40.3306° N, 84.0925° W
ZIP code45360
Census code (GEOID)3964262

Sources: U.S. Census Bureau (Gazetteer, ZCTA), USGS GNIS, Wikidata.

Towns near Port Jefferson

Straight-line distance between town centers.

Frequently asked questions about Port Jefferson

What county is Port Jefferson in?

Port Jefferson is in Shelby County, Ohio.

What is the population of Port Jefferson, OH?

308 people lived in Port Jefferson at the 2020 Census.

What is the ZIP code of Port Jefferson, OH?

The ZIP code of Port Jefferson is 45360.

What is the elevation of Port Jefferson?

971 feet (296 m) above sea level.

What towns are near Port Jefferson?

Sidney (4.9 mi), Anna (6.3 mi), Quincy (7.0 mi), Jackson Center (7.6 mi), De Graff (9.3 mi).

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