what is the entropy change that occurs when 500g of water ice melts at 0 degrees celsius and 1 atm. the heat of fusin for water ice under these conditions is 6.01 Kj/mol.
delta Hfus = Tfus x delta Sfus
Convert 500 g H2O to moles and convert that to kJ, substitute into the above equation and calculate delta Sfusion. Don't forget T must be in Kelvin
To determine the entropy change that occurs when 500g of water ice melts at 0 degrees Celsius and 1 atm, we need to use the formula:
ΔS = ΔH / T
Where:
ΔS - Entropy change
ΔH - Heat of fusion
T - Temperature
First, let's convert the mass of the water ice from grams to moles. The molar mass of water is approximately 18g/mol, so we can calculate the number of moles:
Number of moles = mass / molar mass
Number of moles = 500g / 18g/mol ≈ 27.78 mol
Now, let's calculate the heat of fusion in joules:
Heat of fusion = 6.01 KJ/mol × 1000 J/KJ
Heat of fusion = 6010 J/mol
Next, we'll calculate the entropy change:
ΔS = (Number of moles) × (Heat of fusion) / (temperature in Kelvin)
ΔS = 27.78 mol × 6010 J/mol / 273 K
ΔS ≈ 610 J/K
Therefore, the entropy change that occurs when 500g of water ice melts at 0 degrees Celsius and 1 atm is approximately 610 J/K.