Question

The boiling point of water is lower at higher altitudes. Water boils at 212°F at sea level and 193.7°F at 10,000 ft. What is the percent of decrease in the temperatures, to the nearest tenth of a percent?​

Answer 1

The percent of decrease in the boiling point temperature from sea level to 10,000 ft is 8.6%, calculated by taking the difference in temperatures and dividing by the original temperature at sea level.

Step-by-step explanation:

To calculate the percent of decrease in the boiling point temperature from sea level to 10,000 ft, we first need to find the difference in temperatures in Fahrenheit, and then divide that difference by the original boiling point temperature at sea level. Finally, we multiply by 100 to convert it to a percentage.

The original boiling point at sea level is 212°F, and at 10,000 ft it is 193.7°F. The temperature difference is 212°F - 193.7°F = 18.3°F. The percent decrease is (18.3°F / 212°F) × 100, which is approximately 8.6%.

Therefore, the percent of decrease in the boiling point temperature from sea level to 10,000 ft is 8.6%, to the nearest tenth of a percent.