stem is passed through 40g of ice at -10degree celsius till temperature of mixture becomes 80degree celsius. Mass of mixture will be approximately,
pls ans me.
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.
Explanation:
Given info : stem is passed through 40g of ice at -10°C till temperature of mixture becomes 80°C.
To find : mass of mixture will be approximately ..
solution : let mass of steam is m.
some important data
- specific heat of ice = 0.5 cal/g/°C
- specific heat of water = 1 cal/g/°C
- latent heat of fusion = 80 cal/g
- latent heat of vaporization = 540 cal/g
now, heat lost by steam to change its phase and decrease its temperature upto 80°C , H₁ = mLv + ms(T₂ - T₁)
= m × 540 + m(1)(100 - 80)
= 540 m + 20m
= 560 m
heat gained by ice decrease its temperature , to change its phase and increase its temperature upto 80°C , H₂ = m's'(T₂' - T₁') + m'Lf + m'sT₁
= 40 × 0.5 × 10 + 40 × 80 + 40 × 1 × 80
= 200 + 3200 + 3200
= 6600
from the theory of calorimetry,
heat lost = heat gained
⇒ H₁ = H₂
⇒ 560m = 6600
⇒ m = 6600/560 = 11.7857 g
therefore the mass of steam is 11.7857g and the mass of mixture = (11.7857 + 40)g = 51.7857g