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Question

Determine the final result when 200 gm of water and 20 gm of ice at 0C are in a calorimeter having a water equivalent of 30 gm. 50 gm of steam is passed into it at 100C . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm


A

270 gm ice at 0°C

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B

250 gm ice, 20 gm water at 0°C

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C

270 gm water at 75°C

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D

1 gm steam, 269 gm water at 100°C

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Solution

The correct option is D

1 gm steam, 269 gm water at 100°C


When steam is passed, the final temperature can be 0C , between 0C and 100C , 100C . We will consider all three possibilities

Case I:

Final temperature = 0C

In this case, all the steam condenses and then cools down to 0C.

Heat given out by steam

= 50 × 540 + 50 × 1 × (1000) = 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 0C.

Case II : Final temperature = θ and 0 < θ < 100

Heat lost by steam = heat gained by (ice + water + calorimeter)

50 × 540 + 50 × 1 × (100θ) = 20 × 80 + (20 + 200 + 30)× 1 × (θ0)

θ = 101.3C

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 100C

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 100C

m(540) = 20 × 80 + (20 + 200 + 30) × (1000)

m = 26600540 ≈ 49 gm

49 gm of steam condense and the final temperature is 100C


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