Suppose that the amount of heat removed when 2.9 kg of water freezes at 0.0 oC were removed from ethyl alcohol at its freezing/melting point of -114.4 oC. How many kilograms of ethyl alcohol would freeze?

I don't know where to start with this question! Please help.

To solve this problem, you need to understand the concept of heat transfer and the specific heat capacities of water and ethyl alcohol. Here's how you can approach this question step by step:

1. Determine the heat removed when water freezes:

The heat removed during the freezing process can be calculated using the equation:

Q = m * ΔH_f

Where:
Q is the heat removed (in Joules)
m is the mass of the substance (in kg)
ΔH_f is the specific heat of fusion for the substance (in J/kg)

For water, the specific heat of fusion (ΔH_f) is 334,000 J/kg.

So, for 2.9 kg of water, the heat removed can be calculated as:

Q_water = (2.9 kg) * (334,000 J/kg)

2. Calculate the heat transferred from ethyl alcohol:

Since the heat removed from water is equal to the heat transferred to ethyl alcohol, we can equate the two quantities:

Q_water = Q_ethanol

3. Determine the mass of ethyl alcohol that will freeze:

To find the mass of ethyl alcohol that will freeze, we rearrange the equation from step 2:

m_ethanol = Q_ethanol / ΔH_f_ethanol

The specific heat of fusion for ethyl alcohol (ΔH_f_ethanol) is typically given in the problem or can be found in reference books.

Plug in the values and calculate:

m_ethanol = Q_ethanol / ΔH_f_ethanol

Note: Make sure to convert temperatures to Kelvin (K) before plugging them into any equations. To convert from degrees Celsius (°C) to Kelvin (K), use the equation: T(K) = T(°C) + 273.15.

By following these steps, you should be able to find the answer to the question: how many kilograms of ethyl alcohol would freeze.