Calculating Relative Humidity at 60°F

[jxs919]

My brain is fried on other subjects. Help me!

Given: Max Water Vapor Capacity for Air @ 70 degrees F (21 deg. C)= 8g/cubic foot.

At 60% relative humidity, the actual water vapor content is 4.8g/cubic foot.

If the room temperature were lowered to 60 degrees F (16 deg. C) at which the max water vapor capacity is 5.8g/cubic foot, calculate the new relative humidity.


help me get started here...

 

[wxrocks]

I would start by looking at where 60% (0.60) comes from -- then it will be easy to calculate from there.

 

[Blueshift5]

First, we need to understand the concept of relative humidity. Relative humidity is the amount of water vapor present in the air compared to the maximum amount of water vapor that can be present at a given temperature. It is expressed as a percentage.

In this case, we are given the maximum water vapor capacity for air at 70 degrees F, which is 8g/cubic foot. We are also given that at 60% relative humidity, the actual water vapor content is 4.8g/cubic foot. This means that at 70 degrees F, the air is holding 4.8g of water vapor, which is 60% of its maximum capacity.

Now, we need to calculate the new relative humidity at 60 degrees F. We are given that at this temperature, the maximum water vapor capacity is 5.8g/cubic foot. This means that at 60 degrees F, the air can hold a maximum of 5.8g of water vapor.

To calculate the new relative humidity, we can use the following formula:

Relative Humidity = (Actual water vapor content / Maximum water vapor capacity) x 100

Substituting the values we have, we get:

Relative Humidity = (4.8g/cubic foot / 5.8g/cubic foot) x 100 = 82.76%

Therefore, at 60 degrees F, the relative humidity would be approximately 82.76%. This means that the air is holding 82.76% of its maximum capacity of water vapor at this temperature.

I hope this helps you with your calculation. Remember to always check your units and make sure they are consistent throughout the calculation. Good luck!