Question

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​

Answer 1

Answer:

I need a thermometer for calculating

Answer 2

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.

I hope it will help you...