# 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.

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}