# How much energy is required to change a 58 g ice cube from ice at -1 C to steam at 101 C? The...

## Question:

How much energy is required to change a 58 g ice cube from ice at -1C to steam at 101C? The specific heat of ice is 2090 J/kg.C and of water 4186 J/kg.C. The latent heat of fusion of water is 3.33*10{eq}^5 {/eq} J/kg, its latent heat of vaporization is 2.26*10{eq}^6 {/eq} J/kg, and the specific heat of steam is 2010 J/kg.C. Answer in units of MJ

## Heating an Ice Cube:

To be able to heat an ice cube, certain amounts of energy must be applied. This energy is a sum of two types of energy demand: first, the energy required to heat the ice to a certain temperature; and second, the energies needed to change the phase of the ice (i.e. ice to water, water to steam).

## Answer and Explanation:

We are given the following values:

- the mass of the ice cube, which we convert to kilograms, {eq}m = 58\text{ g}(\dfrac{1\text{ kg}}{1000\text{ g}}) = 0.058\text{ kg} {/eq};

- the initial temperature of the ice cube, {eq}T_0 = -1^\circ \text{C} {/eq};

- the final temperture that we intend to heat the ice cube to, {eq}T = 101^\circ \text{C} {/eq};

- the specific heat of ice, water, and steam, {eq}c_{_{ice}} = 2090\ \mathrm{J/kg \cdot ^\circ C} \ , \ c_{_{water}} = 4186 \ \mathrm{J/kg \cdot ^\circ C} \ , \ c_{_{water}} = 2010\ \mathrm{J/kg \cdot ^\circ C} {/eq};

- the latent heat of fusion and vaporization, {eq}L_{_{fusion}} = 3.33 \times 10^5\ \mathrm{J/kg} \ , \ L_{_{vaporization}} = 2.26 \times 10^6\ \mathrm{J/kg} {/eq};

The energy required for the ice to become steam at the targeted temperature takes five parts:

- heating ice to the melting point;
- changing phase from ice to water;
- heating water to the boiling point;
- changing phase from water to steam, and;
- heating steam to the target temperature.

Mathematically, this can be stated as:

{eq}Q = Q_{_{T_0 \to mp}} + Q_{_{mp}} + Q_{_{water \to bp}} + Q_{_{bp}} + Q_{_{bp \to T}} \\ Q = mc_{_{ice}}(0^\circ \text{ C} - T_0) + mL_{_{fusion}} + mc_{_{water}}(100^\circ \text{ C} - 0^\circ \text{ C} ) + mL_{_{vaporization}} + mc_{_{steam}}(T - 100^\circ \text{ C} ) \\ Q = m(c_{_{ice}}(0^\circ \text{ C} - T_0) + L_{_{fusion}} + c_{_{water}}(100^\circ \text{ C} - 0^\circ \text{ C} ) + L_{_{vaporization}} + c_{_{steam}}(T - 100^\circ \text{ C} )) {/eq}

From here, we plug in the given values:

{eq}Q = m(c_{_{ice}}(0^\circ \text{ C} - T_0) + L_{_{fusion}} + c_{_{water}}(100^\circ \text{ C} - 0^\circ \text{ C} ) + L_{_{vaporization}} + c_{_{steam}}(T - 100^\circ \text{ C} )) \\ Q = (0.058\text{ kg})((2090\ \mathrm{J/kg \cdot ^\circ C})(0^\circ \text{ C} - (-1^\circ \text{C})) +3.33 \times 10^5\ \mathrm{J/kg} + (4186 \ \mathrm{J/kg \cdot ^\circ C})(100^\circ \text{ C} - 0^\circ \text{ C} ) + 2.26 \times 10^6\ \mathrm{J/kg} + (2010\ \mathrm{J/kg \cdot ^\circ C})(101^\circ \text{ C} - 100^\circ \text{ C} )) = 174910.6\text{ J}\\ Q = 174910.6\text{ J} {/eq}

Finally, we convert this to megajoules:

{eq}Q = 174910.6\text{ J}(\dfrac{1\text{ MJ}}{1000\text{ J}}) = 17.49106\text{ MJ}\\ \boxed{Q = 17.5\text{ MJ}} {/eq}

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