Question

In an industrial process $10 \,kg$ of water per hour is to be heated from $20^{\circ} C$ to $80^{\circ} C$. To do this steam at $150^{\circ} C$ 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 $90^{\circ} C$. How many $kg$ of steam is required per hour? (Specific heat of steam = specific heat of water $=1 \, cal / g ^{\circ} C$, Latent heat of vaporization $=540 \,cal / g$ )

Answer 1

Suppose $m kg$ steam is required per hour Heat is released by steam in following three steps
(i) When $150^{\circ} C$ steam $ \xrightarrow[Q_1]{}100^{\circ} C$ steam
$Q_{1}=m c_{\text {steam }} \Delta \theta=m \times 1(150-100)=50 mcal$
(ii) When $100^{\circ} C$ steam $\xrightarrow[Q_2]{} 100^{\circ} C$ water
$Q_{2}=m L_{v}=m \times 540=540\, mcal$
(iii) When $100^{\circ} C$ water $\xrightarrow[Q_2]{} 90^{\circ} C$ water
$Q_{3}=m c_{W} \Delta \theta=m \times 1 \times(100-90)=10\, mcal$
Hence total heat given by the steam
$Q=Q_{1}+Q_{2}+Q_{3}=600 \,mcal$ .....(i)
Heat taken by $10\, kg$ water
$Q^{\prime} m c_{W} \Delta \theta=10 \times 10^{3} \times 1 \times(80-20)=600 \times 10^{3} cal$
Hence $Q=Q^{\prime} \Rightarrow 600 \,m =600 \times 10^{3}$
$\Rightarrow m=10^{3} g =1\, kg$