Question

100g ice at $0^0$ C is put in water in bucket at ${50}^0$ 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

Given: Mass of ice = 100 gm$\left(at{0}^{o}C\right)$

Temperature of bucket= ${50}^{o}C$

Solution:

We know that $S=\frac{△Q}{T}=\frac{mL}{T}$

Let initial entropy be $S_1$ and final entropy be $S_2$

$S_1=80\times \frac{100}{0+273}$

$\therefore S_1=\frac{8000}{273}cal/k$

Now,

$S_2=80\times \frac{100}{273+50}$

$\therefore S_2=\frac{8000}{323}cal/k$

$\Rightarrow △S=(S_2-S_1)$

$\Rightarrow △S=(\frac{8000}{323}-\frac{8000}{273})$

$\Rightarrow △S=-4.5cal/k$

Hence the correct option is A.