Questions

# 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?(Take specific heat of steam = 1 calorie per g°C, Latent heat of vaporization = 540 cal/g)

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a
1 g
b
1 kg
c
10 g
d
10 kg
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detailed solution

Correct option is B

Suppose m kg of steam is required per hour.Heat is released by steam in following three steps(i) When 150∘C steam ⟶Q1100∘C steam Then Q1=mcSteam Δθ=m×1(150−100)=50 m cal(ii) When 100∘C steam ⟶Q2100∘C water Then Q2=mLV=m×540=540 m cal (iii) When 100∘C water ⟶Q390∘C water Then Q3=mcWΔθ=m×1×(100−90)=10 m calHence, total heat given by the steam,Q=Q1+Q2+Q3=600 m cal Hence Q=Q′⇒600m=10×103×60 ⇒m=103g=1kg

Similar Questions

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40oC. When m gram of ice at -10oC is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20oC. It is known that specific heat capacity of the liquid changes with temperature as where $\mathrm{\theta }$ is temperature in oC. The specific heat capacity of ice, water and the calorimeter remains constant and values are  and latent heat of fusion of ice is ${\mathrm{L}}_{\mathrm{f}}=80{\mathrm{calg}}^{-1}$. Assume no heat loss to the surrounding and calculate the value of m in grams.

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