How much water remains unfrozen after 50.4 KJ of heat have been extracted from 258 g of liquid water initially at 0ºC?

We need to calculate how much water remains unfrozen after removing 50.4 kJ of heat from 258 g of liquid water initially at 0ºC.

Step 1: Understand the given data
Heat removed (Q) = 50.4 kJ = 50,400 J
Mass of liquid water (m) = 258 g
Initial temperature = 0ºC
Latent heat of fusion of ice (L) = 334 J/g
Step 2: Calculate the amount of water that freezes
The heat required to freeze a certain mass of water is given by:

Q = m_frozen × L

Where:

Q = Heat removed (in J)
m_frozen = Mass of water that freezes (in g)
L = Latent heat of fusion of ice (in J/g)
Rearranging the equation to find the mass of frozen water:

m_frozen = Q / L
m_frozen = 50,400 J / 334 J/g
m_frozen ≈ 150.9 g

Step 3: Calculate the amount of water that remains unfrozen
Water remaining = Initial mass - Mass frozen
Water remaining = 258 g - 150.9 g
Water remaining ≈ 107.1 g

Step 4: Final Answer
Approximately 107 g of water remains unfrozen.