Suppose that the amount of heat removed when 42.1 kg of water freezes at 0.0 °C were removed from ethyl alcohol at its freezing/melting point of −114.4 °C. How many kilograms of ethyl alcohol would freeze?
heat removed when freezing 42.1 kg H2O = ?
mass H2O x heat fusion = ? joules
42,100 g x heat fusion = ? joules.
Then mass alcohol x heat fusion = joules.
You know heat fusion alcohol (you have that in the problem or you can look it up) and you have joules, solve for mass alcohol.
To solve this problem, we need to use the concept of specific heat and heat of fusion.
The specific heat of water is approximately 4.18 J/g°C, and its heat of fusion is approximately 334 J/g. The specific heat of ethyl alcohol is approximately 2.44 J/g°C, and its heat of fusion is approximately 105 J/g.
Step 1: Calculate the heat removed from water when it freezes.
The heat removed from water can be calculated using the formula:
Q = m * Hf
Where:
Q = heat removed
m = mass of water
Hf = heat of fusion of water
Q = 42.1 kg * 334 J/g
Q = 14,074.4 kJ
Step 2: Calculate the number of kilograms of ethyl alcohol that would freeze.
The heat removed from ethyl alcohol can be calculated using the formula:
Q = m * Hf
Where:
Q = heat removed
m = mass of ethyl alcohol
Hf = heat of fusion of ethyl alcohol
14,074.4 kJ = m * 105 J/g
m = 14,074.4 kJ / 105 J/g
m ≈ 133.61 kg
Therefore, approximately 133.61 kg of ethyl alcohol would freeze.
To determine how many kilograms of ethyl alcohol would freeze, we need to find the heat released when water freezes and equate it to the heat absorbed when ethyl alcohol freezes.
Let's break down the problem step by step:
1. Find the heat released when 42.1 kg of water freezes at 0.0 °C:
The heat released for water freezing is given by the equation:
Q = m * L
where:
Q is the heat released (in Joules)
m is the mass of water (in kilograms)
L is the latent heat of fusion for water (in Joules per kilogram)
The latent heat of fusion for water is approximately 334,000 J/kg.
Plugging in the values, we get:
Q(water) = 42.1 kg * 334,000 J/kg
2. Equate the heat released when water freezes to the heat absorbed when ethyl alcohol freezes:
The heat absorbed for ethyl alcohol freezing is given by the equation:
Q = m * L
where:
Q is the heat absorbed (in Joules)
m is the mass of ethyl alcohol (unknown)
L is the latent heat of fusion for ethyl alcohol (unknown)
We can set up a proportion using the heat released by water and the heat absorbed by ethyl alcohol:
Q(water) = Q(ethyl alcohol)
m(ethyl alcohol) * L(ethyl alcohol) = Q(water)
Plugging in the values we found earlier, we can solve for m(ethyl alcohol):
m(ethyl alcohol) = Q(water) / L(ethyl alcohol)
Note: For this calculation, we need to find the latent heat of fusion for ethyl alcohol.
The latent heat of fusion for ethyl alcohol is approximately 111,000 J/kg.
Now we can plug in the values and solve for m(ethyl alcohol):
m(ethyl alcohol) = (42.1 kg * 334,000 J/kg) / 111,000 J/kg
m(ethyl alcohol) ≈ 126.6 kg
Therefore, approximately 126.6 kilograms of ethyl alcohol would freeze.