In an industrial process,10, kg of water per | vedclass

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(B) Let $m$ be the mass of steam required in $kg$ per hour.Heat gained by $10\, kg$ of water:$Q_{	ext{gain}} = m_w c_w \Delta T = 10\, kg 	imes 1\,

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$1\, kg$

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(B) Let $m$ be the mass of steam required in $kg$ per hour.Heat gained by $10\, kg$ of water:$Q_{	ext{gain}} = m_w c_w \Delta T = 10\, kg 	imes 1\, kcal/kg^\circ C 	imes (80^\circ C - 20^\circ C) = 10 	imes 60 = 600\, kcal = 600,000\, cal$.Heat lost by $m\, kg$ (or $1000m\, g$) of steam:$1$. Cooling steam from $200^\circ C$ to $100^\circ C$: $Q_1 = m_{g} c_{	ext{steam}} \Delta T = (1000m) 	imes 0.5 	imes (200 - 100) = 50,000m\, cal$.$2$. Condensation at $100^\circ C$: $Q_2 = m_{g} L_v = (1000m) 	imes 540 = 540,000m\, cal$.$3$. Cooling water from $100^\circ C$ to $90^\circ C$: $Q_3 = m_{g} c_w \Delta T = (1000m) 	imes 1 	imes (100 - 90) = 10,000m\, cal$.Total heat lost $Q_{	ext{loss}} = Q_1 + Q_2 + Q_3 = (50,000 + 540,000 + 10,000)m = 600,000m\, cal$.Equating heat gained and lost: $600,000m = 600,000 \Rightarrow m = 1\, kg$.

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