Question

5.0 kg of water at 31 ∘C is mixed with 5.0 kg of water at 64 ∘C in a well-insulated container.

Estimate the net change in entropy of the system.

Answer 1

The net change in entropy of the system is 110.989 J/K and is a positive change in entropy.

The system has become more disordered because the two water masses have mixed together to form a uniform mixture, which has a higher entropy than the two separate water masses.

The net change in entropy of the system:

ΔS = m₁c ln(T₂/T₁) + m₂c ln(T₂/T₂)

ΔS = the net change in entropy (J/K)

m₁= the mass of the first water mass (kg)

m₂= the mass of the second water mass (kg)

c = the specific heat capacity of water (J/kg·K)

T₁ = the initial temperature of the first water mass (K)

T₂= the initial temperature of the second water mass (K)

m₁ = m₂ = 5.0 kg

c = 4186 J/kg·K

T₁ = 31 + 273.15 K = 304.15 K

T₂ = 64 + 273.15 K = 337.15 K

Substituting these values into the equation:

ΔS = (5.0 kg)(4186 J/kg·K) ln(337.15 K/304.15 K) + (5.0 kg)(4186 J/kg·K) ln(337.15 K/337.15 K)

ΔS = 110.98907924847617 J/K

ΔS = 110.989 J/K