If 1; g of steam at 100^{circ}C is mixed with | vedclass

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Question

(A) Let us calculate the heat required to convert $1\; g$ of ice at $0^{\circ}C$ to water at $100^{\circ}C$:Heat to melt ice: $Q_1 = m L_f = 1\; g 	i

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Let us calculate the heat required to convert $1\; g$ of ice at $0^{\circ}C$ to water at $100^{\circ}C$:Heat to melt ice: $Q_1 = m L_f = 1\; g 	imes 80\; cal/g = 80\; cal$.Heat to raise water temperature from $0^{\circ}C$ to $100^{\circ}C$: $Q_2 = m c \Delta T = 1\; g 	imes 1\; cal/g^{\circ}C 	imes 100^{\circ}C = 100\; cal$.Total heat required by ice: $Q_{total} = 80 + 100 = 180\; cal$.Now,calculate the heat released by $1\; g$ of steam at $100^{\circ}C$ to condense into water at $100^{\circ}C$:Heat released: $Q_{steam} = m L_v = 1\; g 	imes 540\; cal/g = 540\; cal$.Since the heat released by steam $(540\; cal)$ is greater than the heat required by ice $(180\; cal)$,the final mixture will be at $100^{\circ}C$ with some steam remaining condensed as water.

If $1\; g$ of steam at $100^{\circ}C$ is mixed with $1\; g$ of ice at $0^{\circ}C$,then the resultant temperature of the mixture is ........ $^{\circ}C$.

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