An 11.2 gram block of ice melts at 0'c. what is the change in entropy (in j/k) associated with this process? (the heat of fusion for water ah fus = 6.010 kj/mol.) j/k

ΔS = 1.37 × 10⁻² kJ/K

Explanation:

The change in entropy (ΔS) for the melting process can be calculated using the following expression.

ΔS = Q/T   [1]

where,

Q is the heat absorbed by the block of ice

T is the absolute temperature (in this case 0°C = 273.15K)

Q can be calculated like:

Q = ΔHfus . n = ΔHfus . m / M

where,

ΔHfus is the enthalpy of fusion for water

n is the number of moles

m is the mass

M is the molar mass

Then,

We can replace this value in [1].