# Calculate the amount of heat (in kJ) that must be absorbed to convert 108 g of ice at 0^@"C" to water at 70^@"C"?

Jun 6, 2015

You'd need 67.6 kJ of heat to convert that much ice at ${0}^{\circ} \text{C}$ to water at ${70}^{\circ} \text{C}$.

So, you need to go from ice at ${0}^{\circ} \text{C}$, which is still a solid, to water at 70^@C", which is of course a liquid. This implies that you will go through a phase change, i.e. from ice at ${0}^{\circ} \text{C}$ to water at ${0}^{\circ} \text{C}$.

As a result, you're going to have to consider two heats, one needed for the phase change and the other needed to raise the temperature of water from $0$ to ${70}^{\circ} \text{C}$.

The heat needed for the phase change will be

${q}_{1} = m \cdot \Delta {H}_{\text{fus}}$, where

$m$ - the mass of ice/water;
$\Delta {H}_{\text{fus}}$ - the enthalpy of fusion.

Plug in your values to get

q_1 = 108cancel("g") * 333"J"/cancel("g") = "35964 J"

The heat needed to raise the temperature of the water will be

${q}_{2} = m \cdot {c}_{\text{water}} \cdot \Delta T$, where

${c}_{\text{water}}$ - the specific heat of water;
$\Delta T$ - the change in temperature, defined as the difference between the initial and the final temperature of the water.

Plug in your values to get

q_2 = 108cancel("g") * 4.18"J"/(cancel("g") ^@cancel("C")) * (70-0)^@cancel("C") = "31600.8 J"

The total heat needed will be

${q}_{\text{total}} = {q}_{1} + {q}_{2}$

${q}_{\text{total" = 35964 + 31600.8 = "67654.8 J}}$

I'll leave the answer rounded to three sig figs and expressed in kJ, so you'll get

q_"total" = color(green)("67.6 kJ")