A space 100 Sqm and ceilings 3m high is to be cooled

[x0Fermion0x]

Homework Statement

A space 100 Sqm and ceilings 3m high is to be cooled by an evaporative air conditioner. Calculate the amount of water it takes to cool the space from 30° to 20°. assume that the air in the space is not being heated at the same time it's cooled.

Homework Equations

H2O(l) → H2O(g) = 44.5 kJ/m
q = C x ▲T
q = m x c x ▲T

The Attempt at a Solution

I have calculated q of water(20*C) -> water (100*C) and
q of water vaporization and added the 2 together to get the total q of the water in the cooler.
I have also calculated the q for the air (30*C) -> (20*C)
but I am stuck on how to relate the two together.
I am guessing that as the energy is transferred from the air to the water and since energy is conserved that the energy lost by the air= the energy gained by the water but am still stuck.

 

[nate808]

Thank you for your post. I would like to help you solve this problem. First, let's define some variables:

V = volume of the space (100 sqm x 3m = 300 cubic meters)
T1 = initial temperature of the air (30°C)
T2 = final temperature of the air (20°C)
m = mass of water needed (unknown)
c = specific heat capacity of water (4.186 J/g°C)
L = latent heat of vaporization of water (44.5 kJ/mol)

Now, let's start by calculating the energy needed to cool the air in the space from 30°C to 20°C. We can use the equation q = m x c x ▲T, where q is the energy needed, m is the mass of the air, c is the specific heat capacity of air (1.005 J/g°C), and ▲T is the change in temperature.

q = m x c x (T2 - T1)
q = m x 1.005 x (20 - 30)
q = -10.05 m

Next, we need to calculate the energy needed to convert the water from liquid to gas. We can use the equation q = m x L, where q is the energy needed, m is the mass of water, and L is the latent heat of vaporization of water.

q = m x L
q = m x 44.5

Since we know that the energy lost by the air is equal to the energy gained by the water, we can set these two equations equal to each other and solve for m.

-10.05 m = m x 44.5
-10.05 = 44.5
m = -22.2 g

Therefore, to cool the space from 30°C to 20°C, we will need 22.2 grams of water. Please note that this calculation assumes that all the water will evaporate and that there are no other factors (such as humidity or heat transfer to the surroundings) affecting the cooling process.

I hope this helps. Let me know if you have any further questions or if you need any clarification.