Answer:
Approximately [tex]6.5 \times 10^{3}\; {\rm kJ}[/tex] (or equivalently [tex]6.5 \times 10^{6}\; {\rm J}[/tex]) assuming a melting point of [tex]0\; {\rm ^{\circ} C}[/tex] and a boiling point of [tex]100\; {\rm ^{\circ} C}[/tex].
Explanation:
The total amount of energy required is the sum of five parts:
The specific heat of ice is [tex]c(\text{ice}) = 2.09 \; {\rm kJ \cdot K^{-1} \cdot kg^{-1}}[/tex]. Energy required to raised the temperature of [tex]m = 2.0\; {\rm kg}[/tex] of ice by [tex]\Delta T = (0 - (- 18)) \; {\rm K} = 18\; {\rm K}[/tex]:
[tex]\begin{aligned}Q &= c(\text{ice}) \, m\, \Delta T \approx 75.24\; {\rm kJ} \end{aligned}[/tex].
The enthalpy of fusion (melting) of water is approximately [tex]\Delta H = 334\; {\rm kJ \cdot kg^{-1}}[/tex]. The energy required to melt [tex]m = 2.0\; {\rm kg}[/tex] of ice would be:
[tex]Q = m\, \Delta H \approx 668\; {\rm kJ}[/tex].
The specific heat of water is [tex]c(\text{water}) =4.18 \; {\rm kJ \cdot K^{-1} \cdot kg^{-1}}[/tex]. Energy required to raised the temperature of [tex]m = 2.0\; {\rm kg}[/tex] of water by [tex]\Delta T = (100 - 0) \; {\rm K} = 100\; {\rm K}[/tex]:
[tex]\begin{aligned}Q &= c(\text{water}) \, m\, \Delta T \approx836\; {\rm kJ} \end{aligned}[/tex].
The enthalpy of vaporization of water is approximately [tex]\Delta H = 2257\; {\rm kJ \cdot kg^{-1}}[/tex]. The energy required to vaporize [tex]m = 2.0\; {\rm kg}[/tex] of water would be:
[tex]Q = m\, \Delta H \approx 4.51 \times 10^{3}\; {\rm kJ}[/tex].
The specific heat of steam is [tex]c(\text{water}) =2.09\; {\rm kJ \cdot K^{-1} \cdot kg^{-1}}[/tex]. Energy required to raised the temperature of [tex]m = 2.0\; {\rm kg}[/tex] of steam by [tex]\Delta T = (220 - 100) \; {\rm K} = 120\; {\rm K}[/tex]:
[tex]\begin{aligned}Q &= c(\text{steam}) \, m\, \Delta T \approx 920\; {\rm kJ} \end{aligned}[/tex].
Hence, the total amount of energy required is approximately [tex]6.5 \times 10^{3}\; {\rm kJ}[/tex].
