Questions

In an industrial process 10 kg of water per hour is to be heated from 20^{o}C to 80^{o}C. To do this steam at 150^{o}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^{o}C. How many kg of steam is required per hour?

(Specific heat of steam = 1 calorie per gm^{o}C, 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

NEW

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 40^{o}C. When m gram of ice at -10^{o}C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^{o}C. It is known that specific heat capacity of the liquid changes with temperature as $\mathrm{S}=\left(1+\frac{\mathrm{\theta}}{500}\right)\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$where $\mathrm{\theta}$ is temperature in ^{o}C. The specific heat capacity of ice, water and the calorimeter remains constant and values are ${\mathrm{S}}_{\text{ice}}=0.5\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$; ${\mathrm{S}}_{\text{water}}=1.0\text{cal}{{\mathrm{g}}^{-1}}^{\circ}{\mathrm{C}}^{-1}$ 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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