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(A) The process involves several steps to convert ice at $-10^{\circ} C$ to steam at $110^{\circ} C$:$1$. Heating ice from $-10^{\circ} C$ to $0^{\cir

Vedclass · Accepted Answer

$29.192$

Vedclass · Answer

(A) The process involves several steps to convert ice at $-10^{\circ} C$ to steam at $110^{\circ} C$:$1$. Heating ice from $-10^{\circ} C$ to $0^{\circ} C$: $Q_1 = m \cdot c_{	ext{ice}} \cdot \Delta T = 0.04 \ kg 	imes 0.5 \ kcal/kg^{\circ} C 	imes 10^{\circ} C = 0.2 \ kcal$.$2$. Melting ice at $0^{\circ} C$ to water at $0^{\circ} C$: $Q_2 = m \cdot L_{	ext{fusion}} = 0.04 \ kg 	imes 80 \ kcal/kg = 3.2 \ kcal$.$3$. Heating water from $0^{\circ} C$ to $100^{\circ} C$: $Q_3 = m \cdot c_{	ext{water}} \cdot \Delta T = 0.04 \ kg 	imes 1 \ kcal/kg^{\circ} C 	imes 100^{\circ} C = 4.0 \ kcal$.$4$. Vaporizing water at $100^{\circ} C$ to steam at $100^{\circ} C$: $Q_4 = m \cdot L_{	ext{vap}} = 0.04 \ kg 	imes 540 \ kcal/kg = 21.6 \ kcal$.$5$. Heating steam from $100^{\circ} C$ to $110^{\circ} C$: $Q_5 = m \cdot c_{	ext{steam}} \cdot \Delta T = 0.04 \ kg 	imes 0.48 \ kcal/kg^{\circ} C 	imes 10^{\circ} C = 0.192 \ kcal$.Total thermal energy $Q = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 = 0.2 + 3.2 + 4.0 + 21.6 + 0.192 = 29.192 \ kcal$.

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