wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In an industrial process 10 kg of water per hour is to be heated from 20C to 80C. To do this steam at 150C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90C. then___ kg of steam is required per hour.
(Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporization = 540 cal/gm)

Open in App
Solution

Heat required by 10 kg water to change its temperature from 20C to 80C in one hour is Q1=(mcΔT)water=(10×103)×1×(8020)=600×103 calorie
In condensation (i) Steam release heat when it looses it's temperature from 150C to 100C.
[mcsteamΔT]
(ii) At 100C it converts into water and gives the latent heat. [mL]
(iii) Water release heat when it looses it's temperature from 100C to 90C.
[mcsteamΔT]
If m gm steam condensed per hour, then heat released by steam in converting water of 90C
Q2=(mcΔT)steam+mLsteam+(msΔT)water=m[1×(150100)+540+1×(10090)]=600 m calorie
According to problem Q1=Q2 = m=103 gm=1 kg=1000 grams.


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Hormones in Pregnancy
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon