M kg of water at t^{circ} C is divided into | vedclass

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(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.Heat released by $m \ kg$ of water at $t^{\circ} C$ to become i

Vedclass · Accepted Answer

$\frac{640 - t}{720}$

Vedclass · Answer

(A) Let the mass of the first part be $m$ and the mass of the second part be $(M - m)$.Heat released by $m \ kg$ of water at $t^{\circ} C$ to become ice at $0^{\circ} C$ is:$Q_1 = m 	imes c 	imes (t - 0) + m 	imes L_f = m 	imes 1 	imes t + m 	imes 80 = m(t + 80)$.Heat absorbed by $(M - m) \ kg$ of water at $t^{\circ} C$ to become steam at $100^{\circ} C$ is:$Q_2 = (M - m) 	imes c 	imes (100 - t) + (M - m) 	imes L_v = (M - m) 	imes 1 	imes (100 - t) + (M - m) 	imes 540 = (M - m)(640 - t)$.Equating $Q_1 = Q_2$:$m(t + 80) = (M - m)(640 - t)$$mt + 80m = 640M - Mt - 640m + mt$$80m + 640m = 640M - Mt$$720m = M(640 - t)$$\frac{m}{M} = \frac{640 - t}{720}$.

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