# How much heat energy, in kilojoules, is required to convert 48.0 g of ice at -18.0 degree C to...

## Question:

How much heat energy, in kilojoules, is required to convert {eq}48.0 \ g {/eq} of ice at {eq}-18.0^\circ \ C {/eq} to water at {eq}25.0^\circ \ C {/eq}?

## Heat Transfer and Temperature Change

The change in temperature is directly proportional to the energy gain or loss, that is defined by an equation {eq}Q = mC \Delta T {/eq}.

where

- {eq}m {/eq} is the mass of the substance

- {eq}C {/eq} is the specific heat of the substance

- {eq}\Delta T {/eq} is the difference between the final and initial temperature of the substance.

Moreover, the change in phase of given substance if also directly associated with the energy given in the equation {eq}Q = mL {/eq}, where {eq}L {/eq} is the latent heat of fusion or vaporization of the substance.

## Answer and Explanation:

No solving for the energy needed to raise the temperature of the ice using the heat equation.The total heat (Q) would be the sum of the following:

- (1) {eq}Q_1 {/eq} - the heat needed to raise the ice from {eq}-18.0^\circ\rm{C} {/eq} to {eq}0.0^\circ\rm{C} {/eq}

- (2) {eq}Q_2 {/eq} - the heat needed to melt the ice into water.

- (3) {eq}Q_3 {/eq} - the heat needed to raise the water from {eq}0.0 ^\circ\rm{C} {/eq} to {eq}25^\circ\rm{C} {/eq}.

Note that the mass of water remains the same regardless of phase.

Let

- {eq}C_{ice} = 2.09 \ \rm{J/g.C} {/eq} be the specific heat of ice,

- {eq}C_{water} = 4.184 \ \rm{J/g.C} {/eq} be the specific heat of water,

- {eq}L_{f} = 334 \ \rm{J/g} {/eq} the latent heat of fusion of water,

Solving Q1 to Q3. That is,

{eq}Q_1 = m_{ice}C_{ice}\Delta T_{ice} \\ \ \ \ \ = (48.0\ g)(2.09 \ J/g. C) (18 \ C) \\ \ \ \ \ = 1805.8\ J {/eq}

{eq}Q_2 = m_{ice \rightarrow water}L_f \\ \ \ \ \ = (48.0\ g)(334 \ J/g) \\ \ \ \ \ = 16032\ J {/eq}

{eq}Q_3 = m_{water}C_{water}\Delta T_{water} \\ \ \ \ \ = (48.0\ g)(4.184 \ J/g.C)(25 \ C) \\ \ \ \ \ = 5020.8\ J {/eq}

Therefore, the total energy required is

{eq}Q_{total} = Q_1 + Q_2 + Q_3 \\ Q_{total} = 1805.8\ J + 16032\ J +5020.8\ J\\ \boxed{Q_{total} = 22.86\ kJ} {/eq}

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