at one atmosphere pure water ice melts at 0.0 C. at 12 atm the melting point is -0.085 C. The density of water is 1g.cm^3 at 0 C, while that of ice is 0.917 g/cm^3. From this information, estimate entropy of melting.
To estimate the entropy of melting, we can use the equation:
ΔS = ΔH / T
where ΔS is the change in entropy, ΔH is the enthalpy change, and T is the temperature.
First, let's calculate the enthalpy change (ΔH) using the given melting points:
ΔH = m * Cp * ΔT
where m is the mass of the substance, Cp is the specific heat capacity, and ΔT is the change in temperature.
For water, ΔT = 0°C - (-0.085°C) = 0.085°C = 0.085 K. Since we have the density of water, we can calculate the mass of water that is being melted:
mass_water = ρ_water * volume_water
where ρ_water is the density of water. Here, we assume that the volume of water and ice are the same.
The density of water at 0°C is given as 1 g/cm^3, so:
mass_water = 1 g/cm^3 * volume_water
For ice, the mass of ice being melted is:
mass_ice = ρ_ice * volume_ice
where ρ_ice is the density of ice. The density of ice is given as 0.917 g/cm^3, so:
mass_ice = 0.917 g/cm^3 * volume_ice
Since the volumes of water and ice are assumed to be the same:
ρ_water * volume_water = ρ_ice * volume_ice
volume_water = volume_ice
Therefore, the enthalpy change for the given mass of water and ice is:
ΔH = (m_water * Cp_water - m_ice * Cp_ice) * ΔT
Now, we can estimate the entropy of melting (ΔS) using the equation ΔS = ΔH / T:
ΔS = ΔH / T
where T is the average of the two melting temperatures (0°C and -0.085°C).
By substituting the values into the equation and performing the calculation, we can obtain an estimate of the entropy of melting.