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(B) Let $m \, kg$ be the mass of steam required per hour.Heat released by steam occurs in three steps:$(i)$ Cooling steam from $150^\circ C$ to $100^\

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$1 \, kg$

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(B) Let $m \, kg$ be the mass of steam required per hour.Heat released by steam occurs in three steps:$(i)$ Cooling steam from $150^\circ C$ to $100^\circ C$:$Q_1 = m 	imes c_{steam} 	imes \Delta 	heta = m 	imes 1 	imes (150 - 100) = 50m \, kcal$(ii) Condensation of steam at $100^\circ C$ to water at $100^\circ C$:$Q_2 = m 	imes L_v = m 	imes 540 = 540m \, kcal$(iii) Cooling water from $100^\circ C$ to $90^\circ C$:$Q_3 = m 	imes c_{water} 	imes \Delta 	heta = m 	imes 1 	imes (100 - 90) = 10m \, kcal$Total heat released by steam $Q = Q_1 + Q_2 + Q_3 = 50m + 540m + 10m = 600m \, kcal$.Heat absorbed by $10 \, kg$ of water to heat from $20^\circ C$ to $80^\circ C$:$Q' = M_{water} 	imes c_{water} 	imes \Delta 	heta = 10 	imes 1 	imes (80 - 20) = 600 \, kcal$.By the principle of calorimetry,Heat lost = Heat gained:$600m = 600$$m = 1 \, kg$.

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