Question

30 g ice at 0°C is mixed with 25 g steam at 100 °C find the final temperature and composition

Answer 1

Answer:

Final temperature $100^\circ$C.

Composition: 15 gm steam, 40 gm water.

Explanation:

Given that

30 g ice at 0°C is mixed with 25 g steam at 100 °C

To find:

Final temperature and composition = ?

Solution:

We know that

Latent heat of steam, L$_v$ = 540 cal/gm

Latent heat of ice = 80 cal/gm

Specific heat of water = 1 cal/$gm ^\circ C$

Heat given by 100 °C steam $\rightarrow$ 100 °C water

$Q_1=mL_v = 25\times 540 = 13500\ cal$

Heat required to melt ice (0°C ice $\rightarrow$ 0 °C water)

$Q_2=mL_f = 30\times 80 = 2400\ cal$

$Q_1 > Q_2$ so the whole ice will be melted to water and water will be processes further towards rise in its temperature.

Heat required to convert water at 0°C  $\rightarrow$ 100 °C water

$Q_3=mC\triangle T = 30 \times 1\times 100 = 3000\ cal$

Total heat still left, E = $Q_3-Q_2-Q_1$ = 13500 - 2400 -3000 = 8100 cal

Now, this heat still left will be used to convert water at 100°C  $\rightarrow$ steam at 100 °C.

Mass of steam(at 100 °C) formed:

$m = \dfrac{E}{L}$

$m = \dfrac{8100}{540} = 15\ gm$

Mass of water (at 100 °C): 30 + 25 - 15 = 40 gm

So, the answer is:

Final temperature $\bold{100^\circ}$C.

Composition: 15 gm steam, 40 gm water.

Answer 2

See the above attachment