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In an industrial process, 10 kg of water per hour is to be heated from $20 ° C and 80 ° C$ . 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?
$[Specific heat of steam = 1 cal g^{- 1} ° C^{- 1}, and latent heat of steam = 540 cal g^{- 1}]$

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Detailed Solution

Heat required by 10kg of water to increase its temperature from $20 to 80 ° C$ in one hour is given by

$Q_{1} = {(m c Δ T)}_{w a t e r} = (10 \times 10^{3}) (1) (80 - 20) = 600 k c a l$

If m gram of steam is condensed per hour, the heat released by steam in converting into water at $90 ° C$ $Q_{2} = m c_{s} (150 - 100) + m L_{v} + m c_{w} (100 - 90)$

$i . e Q_{2} = m (1 \times 50 + 540 + 1 \times 10) = 600 m c a l \{∵ C_{s} = C_{w} = 1 c a \lg^{- 1} ° C\}$

According to given problem $Q_{2} = Q_{1}$ $600 m = 600 \times 10^{3}$ $\Rightarrow m = 1 \times 10^{3} g = 1 k g$

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