防雷中心

Riddle: A storm rains an average of 0.8 inches over an area of 135 square miles. How much latent heat of condensation was produced by the storm in order to produce this mass of rainfall? Final answer will be in Joules. Water density = 1 g/cm^3.






Answer to Riddle: In order to know how much Energy is produced we need to determine the mass of the rain that fell. Density = mass / volume. We know the density of water and we can find the volume of the water. Once we find the volume of the water we can solve the density formula for mass.

Volume of rain will be the area the rain fell on times the depth of the rain. The depth of the rain is 0.8 inches. In order to get the final answer in Joules we need to convert to the metric system for the area the rain fell on. Square miles will need to be converted the square centimeters and the rain depth will need to be in centimeters.

The following conversion factors are need to convert square miles to square centimeters:

1 mile = 5,280 feet
12 inches = 1 foot
1 inch = 2.54 centimeters

The conversion may end up the most challenging aspect of finding the answer. Units must be cancelled properly and if units are squared then the number must be squared also.

First convert 135 square miles to square feet

135 mi^2 * (5,280^2 ft^2 / 1 mi^2) = 3,763,584,000 ft^2

Next convert ft^2 to in^2

3,763,584,000 ft^2 * (12^2 in^2 / 1 ft^2) = 5.41956096 * 10^11 in^2

Now convert in^2 to cm^2 to get it into metric units

5.41956096 * 10^11 in^2 * (2.54^2 cm^2 / 1 in^2) = 3.496483949 * 10^12 cm^2

Since we have the area the rain fell over in square centimeters we can multiply by the rain depth to get the volume of rain in cm^3.

0.8 in (2.54 cm / 1 in) = 2.032 cm

Volume of rain = 3.496483949 * 10^12 cm^2 * 2.032 cm = 7.104855384 * 10^12 cm^3

The density formula can now be used to determine the mass of rain

Density = mass / volume

1 g/cm^3 = mass / 7.104855384 * 10^12 cm^3

mass = 7.104855384 * 10^12 cm^3 * 1 g/cm^3 = 7.104855384 * 10^12 grams

The latent heat of condensation has a known value of 2.5 * 10^6 Joules/kg

convert mass to kg since latent heat has kg in units

(7.104855384 * 10^12 grams) * (1 kg / 1,000 g) = 7.104855384 * 10^9 kg

Total energy is equal to mass of rain times the latent heat of condensation

2.5 * 10^6 Joules/kg * 7.104855384 * 10^9 kg = 1.776213846 * 10^16 Joules

written out and rounded this is 17,700,000,000,000,000 Joules

This is 17.7 quadrillion Joules which is an amazing amount of energy just from the condensation produced from a rain complex.

谜语：暴风雨在 0.8 平方英里的区域内平均降雨量为 135 英寸。如何 风暴产生了大量的冷凝潜热，以产生这种潜热 降雨量？最终答案将以焦耳为单位。水密度 = 1 g/cm^3。






谜语的答案：为了知道产生了多少能量，我们需要确定 落下的雨水。密度=质量/体积。我们知道水的密度和 我们可以找到水的体积。一旦我们找到了水的体积，我们就可以解决 质量的密度公式。

降雨量将是降雨面积乘以降雨深度。深度 降雨量为 0.8 英寸。为了得到焦耳的最终答案，我们需要转换为 降雨区域的公制。平方英里将需要转换 平方厘米和雨深需要以厘米为单位。

将平方英里转换为平方厘米需要以下转换系数：

1 英里 = 5,280 英尺 12 英寸 = 1 英尺

1 英寸 = 2.54 厘米

转换可能最终成为寻找答案最具挑战性的方面。单位必须 正确取消，如果单位是平方，则数字必须 平方也。

首先将 135 平方英里转换为平方英尺

135 mi^2 * （5,280^2 ft^2 / 1 mi^2） = 3,763,584,000 ft^2
下一页 将 ft^2 转换为 in^2
3,763,584,000 ft^2 * （12^2 in^2 / 1 ft^2） = 5.41956096 * 10^11 in^2
现在将 in^2 转换为 cm^2 以将其转换为公制单位

5.41956096 * 10^11 in^2 * （2.54^2 cm^2 / 1 in^2） = 3.496483949 * 10^12 cm^2




由于我们有降雨量以平方厘米为单位的面积，我们可以乘以 通过降雨深度得到以 cm^3 为单位的雨量。

0.8 英寸（2.54 厘米/1 英寸）= 2.032 厘米

雨量 = 3.496483949 * 10^12 cm^2 * 2.032 cm = 7.104855384 * 10^12 cm^3 密度公式现在可用于确定雨

的质量 密度 = 质量 /
体积 1 g/cm^3 = 质量 / 7.104855384 * 10^12 cm^3
质量 = 7.104855384 * 10^12 cm^3 * 1 g/
cm^3


= 7.104855384 * 10^12 克

冷凝潜热的已知值为 2.5 * 10^6 焦耳/千克，将质量转换为千克，因为潜热的单位

为 kg （7.104855384 * 10^12 克） * （1 千克/1,000 克） = 7.104855384 * 10^9 千克 总能量等于雨的质量乘以冷凝
潜热
2.5 * 10^6 焦耳/千克 * 7.104855384 * 10^9 千克



= 1.776213846 * 10^16 焦耳

写出来并四舍五入，这是 17,700,000,000,000,000 焦耳
这是 17.7 万亿焦耳
，这是一个惊人的能量量 由雨水复合物产生的冷凝水。