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(A) The total heat $Q$ required is the sum of heat for four stages:$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot c_{ice} \cdot

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$3045$

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(A) The total heat $Q$ required is the sum of heat for four stages:$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: $Q_1 = m \cdot c_{ice} \cdot \Delta T = 1 \cdot 0.5 \cdot 10 = 5\,cal$.$2$. Melting ice at $0^{\circ}C$: $Q_2 = m \cdot L_{f} = 1 \cdot 80 = 80\,cal$.$3$. Heating water from $0^{\circ}C$ to $100^{\circ}C$: $Q_3 = m \cdot c_{water} \cdot \Delta T = 1 \cdot 1 \cdot 100 = 100\,cal$.$4$. Vaporizing water at $100^{\circ}C$: $Q_4 = m \cdot L_{v} = 1 \cdot 540 = 540\,cal$.Total heat $Q = 5 + 80 + 100 + 540 = 725\,cal$.Converting to Joules: $Q = 725 \cdot 4.2 = 3045\,J$.

How many $J$ of heat energy is required to convert $1\,g$ of ice at $-10^{\circ}C$ into steam at $100^{\circ}C$?

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