When 54 g of ice at -20^{circ} C is mixed wi | vedclass

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(B) Given: $m_{	ext{ice}} = 54 \ g$, $T_{	ext{ice}} = -20^{\circ} C$, $m_{	ext{steam}} = 25 \ g$, $T_{	ext{steam}} = 100^{\circ} C$.Heat required

Vedclass · Accepted Answer

$73 \ g$ water at $100^{\circ} C$ and $6 \ g$ steam at $100^{\circ} C$

Vedclass · Answer

(B) Given: $m_{	ext{ice}} = 54 \ g$, $T_{	ext{ice}} = -20^{\circ} C$, $m_{	ext{steam}} = 25 \ g$, $T_{	ext{steam}} = 100^{\circ} C$.Heat required to bring ice to $0^{\circ} C$: $Q_1 = m_{	ext{ice}} \cdot c_{	ext{ice}} \cdot \Delta T = 54 	imes 2.1 	imes 20 = 2268 \ J$.Heat required to melt the ice at $0^{\circ} C$: $Q_2 = m_{	ext{ice}} \cdot L_f = 54 	imes 334 = 18036 \ J$.Total heat required to convert ice at $-20^{\circ} C$ to water at $0^{\circ} C$: $Q_{	ext{total}} = Q_1 + Q_2 = 2268 + 18036 = 20304 \ J$.Heat released by condensing $25 \ g$ of steam at $100^{\circ} C$ to water at $100^{\circ} C$: $Q_{	ext{condense}} = m_{	ext{steam}} \cdot L_v = 25 	imes 2260 = 56500 \ J$.Since $Q_{	ext{condense}} > Q_{	ext{total}}$, all ice melts and the water temperature rises to $100^{\circ} C$.Heat required to raise the temperature of $54 \ g$ of water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m_{	ext{ice}} \cdot c_{	ext{water}} \cdot \Delta T = 54 	imes 4.2 	imes 100 = 22680 \ J$.Remaining heat available for condensation: $Q_{	ext{rem}} = Q_{	ext{condense}} - (Q_{	ext{total}} + Q_3) = 56500 - (20304 + 22680) = 56500 - 42984 = 13516 \ J$.Mass of steam condensed by this remaining heat: $m_{	ext{condensed}} = Q_{	ext{rem}} / L_v = 13516 / 2260 \approx 6 \ g$.Final mass of water = $54 \ g$ (melted ice) + $(25 - 6) \ g$ (condensed steam) = $73 \ g$.Final mass of steam = $6 \ g$ (remaining steam).Thus, the mixture contains $73 \ g$ of water and $6 \ g$ of steam at $100^{\circ} C$.

When $54 \ g$ of ice at $-20^{\circ} C$ is mixed with $25 \ g$ of steam at $100^{\circ} C$, then the final mixture at thermal equilibrium contains

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