Question

${100}_g$ ice at $0^{o}C$ is put in water in bucket at ${50}^{o}C$. On complete melting change in entropy will be (Assuming no change in temperature of water)

• A. $-4.5$cal/K
• B. $+4.5$cal/K
• C. $+5.4$cal/K
• D. $-5.4$cal/K

Answer 1

The correct option is A $-4.5$cal/K

$\text{Solution}:$

Change of entropy of ice

$S_1=\frac{\Delta Q}{T}=\frac{mL}{T}$

= $\frac{80\times 100}{0+273}$

=$8\times {10}^{3}273cal/K$

Change of entropy of water

$S_2=\frac{\Delta Q}{T}=\frac{mL}{T}$

=$\frac{80\times 100}{273+50}$

=$8\times {10}^{3}323cal/K$

Total change of entropy

$S_2-S_1=\frac{8\times {10}^3}{323}-\frac{8\times {10}^3}{273}$

= - 4.5 cal/K

$\text{Hence A is the correct option}$