Question

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

Answer 1

ΔS = 1.37 × 10⁻² kJ/K

Step-by-step 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,

`Q= \Delta H_(fus) * (m)/(M) =6.010 kJ/mol * (11.2g)/(18.0g/mol) =3.74kJ`

We can replace this value in [1].

`\Delta  S=(Q)/(T) =(3.74kJ)/(273.15K) =1.37 * 10^(-2) kJ/K`