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Question

In an industrial process 10kg of water per hour is to be heated from 20Cto80C. 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.

(Given: Specific heat of steam =1 calorie per g C, latent heat of vaporization =540calg1)

A
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C
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Solution

The correct option is A 1
Heat required by 10kg water to change its temperature from 20Cto80C in one hour is,
Q1=(mcΔT)water
where, m is the mass of water, c is the specific heat of water and ΔT is the change in temperature of the water.
Given, c=1 cal, m=10 kg=10×103 g
Q1=(10×103)×1×(8020)
=600×103calorie
In condensation,
(i) Steam release heat when it looses it's temperature from 150Cto100C, i.e., [mcsteamΔT], where, where, m is the mass of steam, csteam is the specific heat of steam and ΔT is the change in temperature of the steam.
(ii) At 100C, it converts into water and gives the latent heat, i.e., [mL], where, m is the mass of ice and L is the latent heat of vaporization.
(iii) Water release heat when it looses it's temperature from 100Cto90C, i.e., [mcwaterΔT], where, m is the mass of water, cwater is the specific heat of water and ΔT is the change in temperature.
If m gram of steam condenses per hour, then heat released by steam in converting water to 90C,
Q2=(mcΔT)steam+mLsteam+(msΔT)water
=m[1×(150100)+540+1×(10090)]
=600mcalorie
According to problem Q1=Q2600m=600×103m=103 g=1kg

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