Answer to Question #90492 in Molecular Physics | Thermodynamics for paul

Question

Calculate the change in entropy when 12g of water at -10oC is raised to 100oC and convert to steam at 100oC. Assume that specific heat of water cwater = 4190 J·g-1·K-1, heat of vaporization Lv = 2256 kJ/kg,specific heat of ice cice = 2220 J·g-1·K-1, heat of fusion LF = 334 kJ/kg.

Answer 1

Using SI:

$T_1=-10^oC+273=263 K$ - tempersture of starting heating

$T_2=0^oC+273=273 K$ - temperature of ice melting

$T_3=100^oC+273=373 K$ - temperature of converting to steam

$m=0.012 kg$ - mass of water

$c_{water}= 4190 Jkg^{-1}K^{-1}$ - specific heat of water //mistake in statment $g \to kg$

$c_{ice}= 2220 Jkg^{-1}K^{-1}$- specific heat of ice //mistake in statment $g \to kg$

$L_v=2256*10^3 Jkg^{-1}$ - heat of vaporization

$L_f=334*10^3 Jkg^{-1}$ - heat of fusion

For specific heat: $C=c*m$

For phase transition: $\Delta H=L*m$

The process could be divided into 4 parts:

Heatind of ice to 0 degrees Celsius. Change of entropy:

$$\Delta S_1=Cln(T_2/T_1)=c_{ice}mln(T_2/T_1)$$

Melting of the ice at constant temperature. Chenge of entropy:

$$\Delta S_2=\Delta H_f/T_1=L_fm/T_1$$

Heatind of water from 0 to 100 degrees Celsius. Change of entropy:

$$\Delta S_3=Cln(T_3/T_2)=c_{water}mln(T_3/T_2)$$

Vaporization. Change of entropy:

$$\Delta S_4=\Delta H_v/T_3=L_vm/T_3$$

Total change of entropy:

$$\Delta S=\Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4 \implies$$

$$\Delta S=c_{ice}mln(T_2/T_1)+L_fm/T_1+c_{water}mln(T_3/T_2)+L_vm/T_3$$

Using numbers:

$$\Delta S=103.96 J/K$$

Answer: $\Delta S=103.96 J/K$