how many moles of liquid water must freeze to remove 100 kJ of heat?
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Heat removed removed by liquid water to freeze = mLf
Where :
m is the mass of liquid water and Lf the latent fusion of ice.
Latent fusion of ice = 334 joules per gram.
Heat removed = 100000 joules
100000 joules = m × 334joules / gram
m = 100000/334 = 299.40 g
Molar mass of water ( H₂O) : 1 × 2 moles of hydrogen + 16 × 1 mole of oxygen = 18g
Moles = mass / molar mass
299.40 / 18 = 16.63 moles.
We therefore need 16.63 moles of liquid water.
Where :
m is the mass of liquid water and Lf the latent fusion of ice.
Latent fusion of ice = 334 joules per gram.
Heat removed = 100000 joules
100000 joules = m × 334joules / gram
m = 100000/334 = 299.40 g
Molar mass of water ( H₂O) : 1 × 2 moles of hydrogen + 16 × 1 mole of oxygen = 18g
Moles = mass / molar mass
299.40 / 18 = 16.63 moles.
We therefore need 16.63 moles of liquid water.
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