Question

Ice is being prepared from water at ${15}^{\circ }C$ using the process shown below. The final temperature of ice is -${10}^{\circ }C$ and the final temperature of water which is used to cool the condenser is ${30}^{\circ }C$. The minimum work required per kg ice for the process is
$\text{Use}(C_p{)}_{\text{water}}=4.18\text{kJ/kgK;}(C_p{)}_{\text{ice}}=2\text{kJ/kgK};L_{\text{ice}}=334\text{kJ/kg}$

• A. 25.53 kJ
• B. 28.29 kJ
• C. 32.23 kJ
• D. 35.67 kJ

Answer 1

The correct option is C 32.23 kJ

For minimum work process should be reversible.
$\Delta S_{ice}=1\times 4.18\times In\left(\frac{273}{288}\right)-\frac{334}{273}+1\times 2\times In\left(\frac{263}{273}\right)$
$\Delta S_{ice}=1.52kJ/K$
$\Delta S_{gen}=0$
$\Rightarrow \Delta S_{ice}+\Delta S_{water}=0$
So, $\Delta S_{water}=-\Delta S_{ice}=1.52kJ/K$
$M_W\times 4.18\times In\left(\frac{303}{288}\right)=1.52$
$M_W=7.16kg$
$Q_1=m_{ice}{C}_{p\omega }(15-0)+m_{ice}{\times }_{ice}+m_{pice}(0+10)$
$Q_1=1\times 4.18\left(15\right)+1\times 334+1\times 2\times 10=416.7kJ$
$Q_2=M_{\omega }{C}_{p\omega }(30-15)=448.93kJ$
$W_{min}=Q_2-Q_1=32.23kJ$