Determine the final result when 200 gm of water and 20 gm of ice at 0∘C are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at 100∘C . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm
1 gm steam, 269 gm water at 100°C
When steam is passed, the final temperature can be 0∘C , between 0∘C and 100∘C , 100∘C . We will consider all three possibilities
Case I:
Final temperature = 0∘C
In this case, all the steam condenses and then cools down to 0∘C.
Heat given out by steam
= 50 × 540 + 50 × 1 × (100−0) = 32000 cal.
Mass of ice which will melt by this heat = 3200080 = 40 gm
But there are only 20 gm of ice in the calorimeter.
Hence final temperature cannot be 0∘C.
Case II : Final temperature = θ and 0 < θ < 100
Heat lost by steam = heat gained by (ice + water + calorimeter)
⇒ 50 × 540 + 50 × 1 × (100−θ) = 20 × 80 + (20 + 200 + 30)× 1 × (θ−0)
⇒ θ = 101.3∘C
the assumption (0 <θ< 100) is proved to be wrong. Hence final temperature cannot be between 0^{\circ}C and 100{\circ}C
⇒ The final temperature will be 100∘C
Case III:Let m = mass of steam condensed.
Heat lost by steam = Heat gained by ice to melt + Heat gained by (water + water + calorimeter) to reach 100∘C
⇒ m(540) = 20 × 80 + (20 + 200 + 30) × (100−0)
⇒ m = 26600540 ≈ 49 gm
⇒ 49 gm of steam condense and the final temperature is 100∘C