Question

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

Answer 1

let the mass of steam required to raise the temperature of hundred gram of water from 24 ℃ to 90℃ be M gram of steam.
each gram of steam on condensing release 536 Calories of heat this time which condenser is at Eight Hundred degree Celsius and equals to final temperature of 90 ℃.
heat released by in gram of condensers steam and condensate water at hundred degree Celsius to water 90℃ the final temperature of the solution m × a specific heat of water fall of temperature in into m×1 × 10m calories.
total heat released by steam condensing and the cooling to 90 ℃ = 536 M + 10 M= 546 M + Calories of Heat.
heat required to raise the temperature of 100 gram of water at 24℃ + m gram of condensed steam from 24 degree celsius to 90℃ = (100 + 1 ) × 1 ×( 90 degree Celsius - 24℃) = ( 100 + m) into 66 calories
using heat gained = heat lost
(100+m)×60=546;=m;=6600+66m=546m;=486m=6600,or m 6600 /480 =13.75 g of steam.

the amount of steam required is 13.75 gram of steam.

Answer 2

Explanation:

Let mass of steam required =m gm

each gram of steam of condensing release 536 calories of heat steam condense at 100°C

cools finally to 90°C

Heat released by m gm of steam on condensing =536×m calne

Final Temp of solution =m× specific heat of water × fall of temp

=m×1×(100−90)

=m×1×10

=10 m clone

heat released =536 m+10 m

546 m calories of heat

Heat required to raised the temp of 100 gm of water at 24°

C+m gm of condensed steam from 24°

C−90

=(100+m)×1×(90−24)

=(100+m)×66 calories

heat gained = Heat lost

(100+m)×66=546 m

6600+66 m=546 m

600=486 m

m=13.75 g of steam

≃13 gm