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Question

(B) Let $m$ be the mass of steam (in grams) condensed to heat the mixture from $25^{\circ} C$ to $70^{\circ} C$.Heat lost by steam = Heat gained by mi

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) Let $m$ be the mass of steam (in grams) condensed to heat the mixture from $25^{\circ} C$ to $70^{\circ} C$.Heat lost by steam = Heat gained by mixture.Heat lost by steam = (Heat of condensation of steam) + (Heat released by condensed water cooling from $100^{\circ} C$ to $70^{\circ} C$).Heat gained by mixture = (Mass of mixture) $	imes$ (Specific heat of water) $	imes$ (Change in temperature).$m L + m s_w (100 - 70) = M s_w (70 - 25)$.Given $L/s_w = 540^{\circ} C$,$M = 500 \, g$,and $\Delta T = 45^{\circ} C$.$m(540 + 30) = 500 	imes 45$.$m = (500 	imes 45) / 570 \approx 39.47 \, g$.Rate of steam supply = $50 \, g/min = 50/60 \, g/s = 5/6 \, g/s$.Time $t_0 = m / 	ext{rate} = 39.47 / (5/6) \approx 47.36 \, s$.The closest value among the options is $45 \, s$.

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