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

Suppose m kg steam required per hour
Heat released by steam in following three steps

$\text{ (i) When }{150}^{\circ }\mathrm{C}\text{ steam }\underset{{\mathrm{Q}}_1}{⟶}{100}^{\circ }\mathrm{C}\text{ steam }$

      ${\mathrm{Q}}_1={\mathrm{mc}}_{\text{Steam }}\mathrm{\Delta \theta }=\mathrm{m}\times 1(150-100)=50\mathrm{m}\text{ cal }$

$\text{ (ii) When }{100}^{\circ }\mathrm{C}\text{ steam }\underset{{\mathrm{Q}}_2}{⟶}{100}^{\circ }\mathrm{C}\text{ water }$

       ${\mathrm{Q}}_2={\mathrm{mL}}_{\mathrm{V}}=\mathrm{m}\times 540=540\mathrm{m}\text{ cal }$

$\text{ (iii) When }{100}^{\circ }\mathrm{C}\text{ water }\underset{{\mathrm{Q}}_3}{⟶}{90}^{\circ }\mathrm{C}\text{ water }$

       $Q_3=m{c}_W\Delta \theta =m\times 1\times (100-90)=10mcal$

Hence total heat given by the steam $\mathrm{Q}={\mathrm{Q}}_1+{\mathrm{Q}}_2+{\mathrm{Q}}_3$ = 600 mcal  ……..(i)

Heat taken by 10 kg water

   ${\mathrm{Q}}^{\mathrm{\prime }}={\mathrm{mc}}_{\mathrm{W}}\mathrm{\Delta \theta }=10\times {10}^3\times 1\times (80-20)=600\times {10}^3\mathrm{cal}$

$\text{ Hence }\mathrm{Q}={\mathrm{Q}}^{\mathrm{\prime }}\Rightarrow 600\mathrm{m}=600\times {10}^3$

      $\Rightarrow \mathrm{m}={10}^3\mathrm{gm}=1\mathrm{kg}$