Cleveland is a town in Emery County, Utah, in the United States. As of the 2000 census, the town population was 508.
Cleveland is located at 39°20′53″N 110°51′19″W / 39.34806°N 110.85528°W / 39.34806; -110.85528 (39.347921, -110.855178).
According to the United States Census Bureau, the town has a total area of 0.9 square miles (2.3 km²), all of it land. Cleveland is a beautiful place located in a swale of land between the mountains and the desert. It embraces about 480 acres (1.9 km2) and is 7 miles (11 km) northeast of Huntington, 18 miles (29 km) southwest of Price, and 17 miles (27 km) northeast of Castle Dale.