Questions

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

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By Expert Faculty of Sri Chaitanya
a
1 gm
b
1 kg
c
10 gm
d
10 kg
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detailed solution

Correct option is B

Suppose m kg steam required per hourHeat released by steam in following three steps (i) When 150∘C steam ⟶Q1100∘C steam       Q1=mcSteam Δθ=m×1(150−100)=50m cal  (ii) When 100∘C steam ⟶Q2100∘C water        Q2=mLV=m×540=540m cal  (iii) When 100∘C water ⟶Q390∘C water        Q3=mcWΔθ=m×1×(100−90)=10mcalHence total heat given by the steam Q=Q1+Q2+Q3 = 600 mcal  ……..(i)Heat taken by 10 kg water   Q′=mcWΔθ=10×103×1×(80−20)=600×103cal Hence Q=Q′⇒600m=600×103      ⇒m=103gm=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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