steam is passed into 54 gm water at 30 degree Celsius till the temperature of the mixture become 90⁰calcius calculate the mass of the mixture explain it and find the value
Answers
Answer:
Let the mass of steam required to raise the temperature of 54 g of water from 30°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=10m calories.
Total heat released by steam condensing and then cooling to 90°C=536m+10m=546m calories of heat.
Heat required to raise the temperature of 54 g of water at 30°C + m gram of condensed steam from 30°C to 90°C $$= (54 + m) ×1×(90°C - 30°C)= (54 + m)×60 calories
Using heat gained = Heat lost
(54+m)×60 =546m; ==>3240+60m =546m;
====> 486m =3240,
orm=3240/486=80g of steam.
Answer:
In sta gram :- @1203_adi_x ^_^