Question

The total amount of energy it takes to heat 2.0 kg of ice from –18oC to 220oC is

Answer 1

Approximately
`6.5 * 10^(3)\; {\rm kJ}` (or equivalently
`6.5 * 10^(6)\; {\rm J}`) assuming a melting point of
`0\; {\rm ^(\circ) C}` and a boiling point of
`100\; {\rm ^(\circ) C}`.

Step-by-step explanation:

The total amount of energy required is the sum of five parts:

• Energy required to heat ice from
  `(-18\; {\rm ^(\circ) C})` to
  `0\; {\rm ^(\circ) C}`.
• Energy required to melt the ice.
• Energy required to heat water from
  `0\; {\rm ^(\circ) C}` to
  `100\; {\rm ^(\circ) C}`.
• Energy required to vaporize the water.
• Energy required to heat steam from
  `100\; {\rm ^(\circ) C}` to
  `220\; {\rm ^(\circ) C}`.

The specific heat of ice is
`c(\text{ice}) = 2.09 \; {\rm kJ \cdot K^(-1) \cdot kg^(-1)}`. Energy required to raised the temperature of
`m = 2.0\; {\rm kg}` of ice by
`\Delta T = (0 - (- 18)) \; {\rm K} = 18\; {\rm K}`:

`\begin{aligned}Q &= c(\text{ice}) \, m\, \Delta T \approx 75.24\; {\rm kJ} \end{aligned}`.

The enthalpy of fusion (melting) of water is approximately
`\Delta H = 334\; {\rm kJ \cdot kg^(-1)}`. The energy required to melt
`m = 2.0\; {\rm kg}` of ice would be:

`Q = m\, \Delta H \approx 668\; {\rm kJ}`.

The specific heat of water is
`c(\text{water}) =4.18 \; {\rm kJ \cdot K^(-1) \cdot kg^(-1)}`. Energy required to raised the temperature of
`m = 2.0\; {\rm kg}` of water by
`\Delta T = (100 - 0) \; {\rm K} = 100\; {\rm K}`:

`\begin{aligned}Q &= c(\text{water}) \, m\, \Delta T \approx836\; {\rm kJ} \end{aligned}`.

The enthalpy of vaporization of water is approximately
`\Delta H = 2257\; {\rm kJ \cdot kg^(-1)}`. The energy required to vaporize
`m = 2.0\; {\rm kg}` of water would be:

`Q = m\, \Delta H \approx 4.51 * 10^(3)\; {\rm kJ}`.

The specific heat of steam is
`c(\text{water}) =2.09\; {\rm kJ \cdot K^(-1) \cdot kg^(-1)}`. Energy required to raised the temperature of
`m = 2.0\; {\rm kg}` of steam by
`\Delta T = (220 - 100) \; {\rm K} = 120\; {\rm K}`:

`\begin{aligned}Q &= c(\text{steam}) \, m\, \Delta T \approx 920\; {\rm kJ} \end{aligned}`.

Hence, the total amount of energy required is approximately
`6.5 * 10^(3)\; {\rm kJ}`.