Steam at 100,^{circ}C is passed into 22,g of | vedclass

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(A) Let $m$ be the mass of steam condensed into water.Heat lost by steam = Heat gained by water.Heat lost by steam = (Heat released during phase chang

Vedclass · Accepted Answer

$24.8$

Vedclass · Answer

(A) Let $m$ be the mass of steam condensed into water.Heat lost by steam = Heat gained by water.Heat lost by steam = (Heat released during phase change) + (Heat released by condensed water cooling from $100\,^{\circ}C$ to $90\,^{\circ}C$).Heat lost = $m 	imes L_v + m 	imes c_w 	imes \Delta T_1 = m 	imes 540 + m 	imes 1 	imes (100 - 90) = 550m$.Heat gained by $22\,g$ of water to reach $90\,^{\circ}C$ from $20\,^{\circ}C$ = $m_{water} 	imes c_w 	imes \Delta T_2 = 22 	imes 1 	imes (90 - 20) = 22 	imes 70 = 1540\,cal$.Equating heat lost and gained: $550m = 1540$.$m = 1540 / 550 = 2.8\,g$.The total mass of water present = (Initial mass of water) + (Mass of condensed steam) = $22\,g + 2.8\,g = 24.8\,g$.

Steam at $100\,^{\circ}C$ is passed into $22\,g$ of water at $20\,^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90\,^{\circ}C$ (Latent heat of steam is $540\, cal/g$) is ............ $g$.

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