Question

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. What will be the composition of the system when thermal equilibrium is reached? Latent heat of fusion of ice = 3.36 × 103 J kg−1 and latent heat of vaporization of water = 2.26 × 106 J kg−1.

Answer 1

1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C. The composition of the system when thermal equilibrium is reached is 1335 gram of water and 665 gram of Steam at 100°C

Explanation:

Mass of ice = 1 kg

Mass of steam = 1 kg

Heat needed to melt 1kg of ice

$Q=mL$

$Q=1\times 3.36\times 10^5$

$\implies Q=336000$ Joule

The ice converted into water will be at 0°C

Heat needed to increase the temperature of water from 0°C to 100°C

Using

$Q=ms\Delta T$

$Q'=1\times 4200\times (100-0)$   (Specific heat of water = 4200 J/kg°C)

$\implies Q'=420000$ Joule

Total heat required = 420000 + 336000 = 756000 Joule

Let m mass of steam is converted into water at 100°C by giving out this much amount of heat

Then

$m\times 2.26\times 10^6=423360$

$\implies m=\frac{756000}{2.26\times 10^6}$

$\implies m=0.335$ kg

Therefore at equilibrium, the steam will be = 1 - 0.335 = 0.665 kg

And water will be 1 + 0.335 = 1.335 kg

The temperature of both will be 100°C

Thus,

The final composition of the system

Water = 1335 g, Steam = 665 g

Hope this answer is helpful.

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