In an industrial process, 10 kg of water per hour is to be heated from 20⁰C to 80⁰C. To do this, steam at 150⁰C 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 90⁰C, how many kg of steam is required per hour? (Take, specific heat of steam = 1 cal per g⁰C, latent heat of vaporisation = 540 cal g^-1)

Suppose m gm steam is required per hour.

Heat is released by steam in following three steps: 

(i) When steam at  ${150}^{\circ }\mathrm{C}\to \text{ steam at }{100}^{\circ }\mathrm{C}$

    $Q_1=m{c}_{\text{steam }}\mathrm{\Delta }\theta =m\times 1(150-100)=50m\text{ cal }$

(ii) When steam at   ${100}^{\circ }\mathrm{C}\underset{Q_2}{\to }\text{ water at }{100}^{\circ }\mathrm{C}$

      $Q_2=m{L}_V=m\times 540=540m\mathrm{cal}$

(iii) When water at ${100}^{\circ }\mathrm{C}\underset{Q_2}{\to }\text{ water at }{90}^{\circ }\mathrm{C}$

    $Q_3=m{c}_w\mathrm{\Delta }\theta =m\times 1\times (100-90)=10m\mathrm{cal}$

     Hence, total heat given by the steam,

     $Q=Q_1+Q_2+Q_3=600\mathrm{mcal}$               …(i)

    Heat taken by 10 kg water,

           $Q^{\mathrm{\prime }}=m{c}_w\mathrm{\Delta }\theta =10\times {10}^3\times 1\times (80-20)=600\times {10}^3$ cal
     As,   $Q=Q^{\mathrm{\prime }}\Rightarrow 600m=600\times {10}^3\Rightarrow m={10}^{3}g=1\mathrm{kg}$