Question

In an industrial process 10 kg of water per hour is to be heated from 20∘C to 80∘C. To do this steam at 150∘C 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 90∘C. how many kg of steam is required per hour. (Specific heat of steam = 1 calorie per gm°C, Latent heat of vaporization = 540 cal/gm)

A
1 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
1 kg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
10 gm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
10 kg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

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

Suggest Corrections
0
Related Videos
Calorimetry
PHYSICS
Watch in App
Explore more