 # The temperature of 100 gm of water is to be raised from 24 degree Celsius to 90 degree Celsius by adding steam to it.tje mass of the steam required for this purpose is

5 years ago

Let the mass of steam required to raise the temperature of 100 g of water from 24°C to 90°C be m gram of steam.

Each gram of steam on condensing releases 536 calories of heat. The steam which condenses is at 100°C, and it cools to final temperature of 90°C.

Heat released by m gram of steam on condensing= 536 × m calorie

Heat released by m gram of condensed steam condensed to water at 100°C to water at 90°C, the final temperature of the solution= m×specific heat of water× fall of temperature= m×1×10 = 10 m calories.

Total heat released by steam condensing and then cooling to 90°C = 536 m + 10 m = 546 m calories of heat.

Heat required to raise the temperature of 100 g of water at 24°C + m gram of condensed steam from 24°C to 90°C = (100 + m) ×1×(90°C - 24°C)= (100 + m)×66 calories

Using heat gained = Heat lost

(100 + m)×66 = 546 m; ==> 6600 + 66 m = 546 m; ====> 486 m = 6600, or m = 6600/480= 13.75 g of steam.

The amount of steam required is 13.75 g of steam. Remember one gram of water is about 1cc and it expands to about 1640 cc of steam at atmospheric pressure.

2 years ago

the heat energy required for 100grams water to raise temp from 24-degree centigrade to 90 degrees =100(1)(90–24)=6600 calory

the latent heat of steam is 540 per gram of steam therfore amount of stem required is=6600/540=12.22grams