Question

Find the difference in entropy between one gram of ice 0°C and one gram of steam at 100°C. Latent heat of fission of ice is at 80 calories specific heat of water is 1 and latent heat of steam at 100°C with 540 calories

Answer 1

Let the final temperature of the mixture will become T.

$Q_1=1\times 540+1\times 1\times(100-T)$

now energy required to convert ice to water

$Q_2=80+T$

Total energy always be conserve,

So,

$540+100-T=80+T$

$\Rightarrow 2T = 560$

$\Rightarrow T = 280^\circ C$

$\Delta S_1 = \frac{540}{273}$

$\Delta S_2 =-\frac{180}{273}$

$\Delta S_3=\frac{360}{273}$

$\Delta S=0$

Hence total change in entropy =0