Suppose the amount of heat removed when 3.0kg of water freezes at 0 degree C were removed from ethyl alcohol and its freezing/melting point of -114 degree C. How many kilograms of ethyl alcohol would freeze?
PLEASE GIVE ME SOME HINTS!!!THANKS A LOT!!!!
(m_(water ) 〖Lf〗_water)/〖Lf〗_alcohol =m_alcohol
(3.0×(33.5×〖10〗^4 ))/((10.8×〖10〗^4 ) )=m_alcohol=9.3kg
Maryo you are right! :)
Look up the heat of fusion of ethanol.
Heat in water freezing= masswater*Hfwater
Heat in alchol freezing=massalcohol*Hfalc
set the equal, and solve for mass of alcohol.
Am i Correct The correct answer is 9.3 kg ?
i guess the correct answer is 9.3 kg. but i don't know how to get this answer?
Well, I'm here to lighten up the mood and give you some hints with a touch of humor!
1. Start by calculating the heat removed when 3.0kg of water freezes at 0 degrees C. Remember, the heat removed is equal to the heat of fusion (latent heat of fusion) multiplied by the mass of the substance. So, put on your "cool" hat and crunch those numbers.
2. Next, think about the heat being transferred from the water to the ethyl alcohol. This heat transfer can be considered as a sort of "chilly exchange" between the two substances.
3. Now, take into account the freezing/melting point of ethyl alcohol at -114 degrees C. This means that the ethyl alcohol needs to be cooled all the way down to -114 degrees C in order to freeze.
4. Using the heat removed from the 3.0kg of water, figure out how much ethyl alcohol can be cooled to -114 degrees C. This will require a bit of "cool" math, but I believe in you!
Remember, it's just a "cool" puzzle to solve. Don't let it freeze your brain!
To find the number of kilograms of ethyl alcohol that would freeze, we need to use the concept of heat transfer during phase change. The heat transferred during a phase change is given by the equation:
q = m * L
where q is the heat transferred, m is the mass of the substance, and L is the latent heat of fusion.
In this case, we know:
- Mass of water frozen, m_water = 3.0 kg
- Latent heat of fusion for water, L_water = 334,000 J/kg
- Latent heat of fusion for ethyl alcohol is assumed to be similar to water, so L_ethanol = 334,000 J/kg
First, let's calculate the heat transferred when the water freezes:
q_water = m_water * L_water
Next, we can equate the heat transferred during the water freezing to the heat that needs to be transferred for the ethyl alcohol to freeze:
q_water = q_ethanol
m_water * L_water = m_ethanol * L_ethanol
Now we can solve this equation to find the mass of ethyl alcohol that would freeze:
m_ethanol = (m_water * L_water) / L_ethanol
Plug in the known values to calculate the mass of ethyl alcohol that would freeze.