Suppose that the amount of heat removed when 13.8 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?
Two problems here wrapped up in one. First you must determine how much heat was removed from the water. That's
q = mass H2O x heat fusion
q = 13,800 g H2O x heat fusion = ?
You must look up heat fusion.
Next, use q = mass EtOH x heat fusion for EtOH. Plug in heat fusion and q and solve for mass EtOH.
Post your work if you get stuck.
To solve this problem, we need to use the heat equation:
Q = m * L
where:
Q is the amount of heat transferred
m is the mass of the substance
L is the latent heat of fusion for the substance
For water, the latent heat of fusion is 333.55 kJ/kg, while for ethyl alcohol, it is 112.51 kJ/kg.
Let's start by calculating the heat transferred when 13.8 kg of water freezes:
Q_water = m_water * L_water
= 13.8 kg * 333.55 kJ/kg
Next, let's calculate the mass of ethyl alcohol that would freeze:
Q_water = Q_ethanol
13.8 kg * 333.55 kJ/kg = m_ethanol * 112.51 kJ/kg
Now, we can solve for the mass of ethyl alcohol:
m_ethanol = (13.8 kg * 333.55 kJ/kg) / 112.51 kJ/kg
Calculating this, we find:
m_ethanol ≈ 41.0 kg
Therefore, approximately 41.0 kg of ethyl alcohol would freeze.