1 text{ kg} of steam at 150^{circ} text{C} is | vedclass

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(C) Mass of steam,$m_s = 1 	ext{ kg} = 1000 	ext{ g} = 10^3 	ext{ g}$.Initial temperature of steam,$T_i = 150^{\circ} 	ext{C}$.Final temperature o

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(C) Mass of steam,$m_s = 1 	ext{ kg} = 1000 	ext{ g} = 10^3 	ext{ g}$.Initial temperature of steam,$T_i = 150^{\circ} 	ext{C}$.Final temperature of steam (as water),$T_f = 90^{\circ} 	ext{C}$.Latent heat of steam,$L_v = 540 	ext{ cal g}^{-1}$.Specific heat of steam,$c_s = 1 	ext{ cal g}^{-1} {}^{\circ} 	ext{C}^{-1}$.Specific heat of water,$c_w = 1 	ext{ cal g}^{-1} {}^{\circ} 	ext{C}^{-1}$.Heat lost by steam $(Q_{lost})$ consists of cooling steam from $150^{\circ} 	ext{C}$ to $100^{\circ} 	ext{C}$,condensation at $100^{\circ} 	ext{C}$,and cooling water from $100^{\circ} 	ext{C}$ to $90^{\circ} 	ext{C}$.$Q_{lost} = m_s c_s (150-100) + m_s L_v + m_s c_w (100-90)$$Q_{lost} = 1000 	imes 1 	imes 50 + 1000 	imes 540 + 1000 	imes 1 	imes 10$$Q_{lost} = 50,000 + 540,000 + 10,000 = 600,000 	ext{ cal} = 60 	imes 10^4 	ext{ cal}$.Heat gained by $20 	ext{ L}$ of water $(Q_{gained})$: $m_w = 20 	ext{ kg} = 20,000 	ext{ g}$.$Q_{gained} = m_w c_w \Delta T = 20,000 	imes 1 	imes \Delta T = 20,000 \Delta T$.By the principle of calorimetry,$Q_{lost} = Q_{gained}$.$600,000 = 20,000 \Delta T$.$\Delta T = \frac{600,000}{20,000} = 30^{\circ} 	ext{C}$.

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