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In an industrial process $10\, kg$ of water per hour is to be heated from $20°C$ to $80°C$. To do this steam at $150°C$ 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 $90°C.$ how many $kg$ of steam is required per hour. $($Specific heat of steam $= 1$ $calorie \,per\, gm°C,$ Latent heat of vaporisation $= 540 \,cal/gm)$
$1 \,gm$
$1\, kg$
$10\, gm$
$10 \,kg$
Solution
(b) Suppose $m\, kg$ steam required per hour
Heat released by steam in following three steps
$(i)$ When $150°C$ steam $\xrightarrow[{{Q_1}}]{}100^\circ C$ steam
$Q_1 = mc_{Steam} \Delta \theta = m \times 1 (150 -100) = 50 m cal$
$(ii)$ When $150°C$ steam $\xrightarrow[{{Q_2}}]{}100^\circ C$ water
$Q_2 = mL_V = m \times 540 = 540 m cal$
$(iii)$ When $100°C$ water $\xrightarrow[{{Q_2}}]{}90^\circ C$ water
$Q_3 = mc_W \Delta \theta = m \times 1 \times (100 -90) = 10 m cal$
Hence total heat given by the steam $Q = Q_1 +Q_2 + Q_3$ = 600 mcal … $(i)$
Heat taken by $10 \,kg$ water
$Q' = m{c_W}\Delta \theta = 10 \times {10^3} \times 1 \times (80 – 20) = 600 \times {10^3}cal$
Hence $Q = Q' ==> 600 m = 600 \times 10^3$
$==> m = 103 gm = 1kg.$
