The heat of vaporization of ethyl alcohol is about 200 cal/g.If 2.4 kg of this fluid were allowed to vaporize in a refrigerator, find what mass of ice would be formed from 0 degree C water.
for the first equation, where do you get the 1000 g/kg from?
200 cal/g * 1000g/kg * 2.4kg = 480000 cal
The latent heat required to turn 0C water to 0C ice is 80cal/g
480000 / 80 = 6000g or 6.0kg
1000 grams per 1 Kilogram, just a way to get correct units
Well, well, well, it seems we have a chemistry question here. Let's see if I can add some fun to it!
So, we have 2.4 kg of ethyl alcohol that wants to vaporize in a fridge. First thing's first, we need to figure out how much heat is needed for that.
Given the heat of vaporization of ethyl alcohol is about 200 cal/g, we can use that to calculate the total amount of heat needed.
2.4 kg = 2400 g (converting to grams)
Total heat = mass x heat of vaporization
Total heat = 2400 g x 200 cal/g
Total heat = 480,000 cal
Now, let's see how much ice would form from the water at 0 degrees Celsius.
To find that out, we need to know how much heat is released when water freezes. The heat of fusion for water is about 80 cal/g.
Using that, we can calculate the mass of ice formed.
Mass of ice = Total heat released / Heat of fusion
Mass of ice = 480,000 cal / 80 cal/g
Mass of ice = 6,000 g
So, my friend, approximately 6,000 grams (or 6 kilograms) of ice would be formed from that freezing water. Enjoy your frosty surprise!
To find the mass of ice formed, we need to calculate the amount of heat removed from the water to freeze it, and then convert it to mass using the heat of fusion of ice.
To calculate the heat removed from the water, we can use the equation:
Q = m * H
where Q is the heat in calories, m is the mass in grams, and H is the specific heat capacity in cal/g°C.
Given that the initial temperature of the water is 0°C, the final temperature of the ice is 0°C (since it is freezing), and the heat capacity of water is 1 cal/g°C, the equation becomes:
Q = m * H * ΔT
Since the temperature change (ΔT) is 0, the heat removed (Q) is also 0.
Since the heat removed is equal to the heat gained by the ethyl alcohol evaporating, we can use the equation:
Q = m * H_vap
where m is the mass in grams and H_vap is the heat of vaporization in cal/g.
Using the given heat of vaporization of 200 cal/g and the mass of the ethyl alcohol, which is 2.4 kg (convert to grams), we can calculate the heat removed:
Q = m * H_vap
Q = 2400 g * 200 cal/g
Q = 480,000 cal
Since the heat removed from the water is 0, the heat gained from the ethyl alcohol evaporation should also be 0. However, this is not possible since the ethyl alcohol is removing heat from the surroundings (refrigerator).
To balance the heat equation, we assume that the heat gained by the ethyl alcohol is equal to the heat removed from the water, which is 480,000 cal. Therefore:
Q_ice = Q_ethyl_alcohol
m_ice * H_fusion = m_ethyl_alcohol * H_vap
We know that the heat of fusion of ice is 80 cal/g. Rearranging the equation:
m_ice = (m_ethyl_alcohol * H_vap) / H_fusion
m_ice = (2400 g * 200 cal/g) / 80 cal/g
m_ice = 6000 g
Therefore, approximately 6000 grams or 6 kg of ice would be formed from 0°C water.