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Harcourt, Iowa

Harcourt, Iowa
City
Location of Harcourt, Iowa
Location of Harcourt, Iowa
Coordinates: 42°15′43″N 94°10′36″W / 42.26194°N 94.17667°W / 42.26194; -94.17667Coordinates: 42°15′43″N 94°10′36″W / 42.26194°N 94.17667°W / 42.26194; -94.17667
Country  United States
State  Iowa
County Webster
Area
 • Total 1.00 sq mi (2.59 km2)
 • Land 1.00 sq mi (2.59 km2)
 • Water 0 sq mi (0 km2)
Elevation 1,175 ft (358 m)
Population (2010)
 • Total 303
 • Estimate (2012) 297
 • Density 303.0/sq mi (117.0/km2)
Time zone Central (CST) (UTC-6)
 • Summer (DST) CDT (UTC-5)
ZIP code 50544
Area code(s) 515
FIPS code 19-34410
GNIS feature ID 0457242

Harcourt is a city in Webster County, Iowa, United States. The population was 303 at the 2010 census.

Harcourt was platted in 1881. It was named for William Vernon Harcourt, a British statesman. A post office has been in operation in Harcourt since 1882.

Harcourt is located at 42°15′43″N 94°10′36″W / 42.26194°N 94.17667°W / 42.26194; -94.17667 (42.262056, -94.176589).

According to the United States Census Bureau, the city has a total area of 1.00 square mile (2.59 km2), all of it land.

As of the census of 2010, there were 303 people, 131 households, and 85 families residing in the city. The population density was 303.0 inhabitants per square mile (117.0/km2). There were 146 housing units at an average density of 146.0 per square mile (56.4/km2). The racial makeup of the city was 98.0% White, 0.3% African American, 0.3% Native American, 0.3% Asian, and 1.0% from two or more races. Hispanic or Latino of any race were 1.3% of the population.


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