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 vaporization = 540 cal/gm)
Text Solution
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Suppose m kg steam required per hour
Heat released by steam in following three steps
$(i) When 150°C steam \xrightarrow{Q_1} 100°C steam Q_1 = mc_{\text{steam}} \Delta \theta = m \times 1 (150 - 100) = 50 m \text{ cal} (ii) When 150°C steam \xrightarrow{Q_2} 100°C water Q_2 = mL_v = m \times 540 = 540 m \text{ cal} (iii) When 100°C water \xrightarrow{Q_2} 90°C water$
Q 3 = mc W Δ Δ θ θ = m × × 1 × × (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
$Q' = m c_W \Delta \theta = 10 \times 10^3 \times 1 \times (80 - 20) = 600 \times 10^3 \text{ cal}$
Hence Q = Q ′ ′ ⇒ ⇒ 600 m = 600 × × 10 3
⇒ ⇒ m = 10 3 gm = 1kg.
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