Question

If 64000 calories of heat are extracted from 100 gm of steam at 100°C, what will be (i) the final temperature

please help me want correct answer please help me​

Answer 1

Answer:

Explanation:

At 100°C, the whole water would have been converted into steam

The heat required to bring the steam at 100°C to water at 100°C

Q₁ = m × Latent Heat of vapourisation

 = 100 × 540

= 54000 calories

Again lets calculate how much heat is required to bring the temperature of water at 100°C to water at 0°C

If the sum of both the heats is less greater than 64800 then the water sill remain at some temperature above 0°C

Using

Q₂ = 100 × 1 × (100 - 0)

      = 10000 calories

Q₁ + Q₂ = 54000 + 10000 = 64000 calories

But the total heat that was extracted was 64800 calories

So still 64800 - 64000 calories = 800 calories of heat is left

Lets assume that m gram of water at 0°C is converted into ice at 0°C with this 800 calories

then

Q = m × Latent Heat of fusion of ice  

800 = m × 80

⇒ m =10 gram

Therefore at at 0°C, 10 gram of water will have converted into 10 gram of ice and 90 gram of water will remain at 0°C