Calculate the entropy change for a process in which 3.00 moles of liquid water at 08c is mixed with 1.00 mole of water at 100.8c in a perfectly insulated container. (assume that the molar heat capacity of water is constant at 75.3 j k21 mol21.)

The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:

ΔS = nCln (T₂/T₁)

n = number of moles

C = molar heat capacity

T₂ = final temperature

T₁ = initial temperature

We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:

[ (3 moles) (273.15 K) + (1 mole) (373.15 K) ] / (4 moles) = 298.15 K

Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:

ΔS = (3 moles) (75.3 J/Kmol) ln (298.15/273.15)

ΔS = 19.8 J/K

ΔS = (1 mole) (75.3 J/Kmol) ln (298.15/373.15)

ΔS = - 16.9 J/K

Now we combine the entropy change of each portion of water to get the total entropy change for the system:

ΔS = 19.8 J/K + (-16.9 J/K)

ΔS = 2.9 J/K

The entropy change for combining the two temperatures of water is 2.9 J/K.