# Question da803

Aug 9, 2015

You need 15.8 kJ.

#### Explanation:

The idea here is that you need to take into account the heat that you must supply to the ice at ${0}^{\circ} \text{C}$ to get it to water at ${0}^{\circ} \text{C}$, then to the water at ${0}^{\circ} \text{C}$ to get it water at ${25}^{\circ} \text{C}$.

An important thing to notice here is that the enthalpy of fusion is given to you in kilojoules per mole. You can convert this to kJ per gram to make the rest of the calculations easier by using water's molar mass

$6.01 \text{kJ"/color(red)(cancel(color(black)("mol"))) * (1color(red)(cancel(color(black)("mol"))))/("18.02 g") = "0.335 kJ/g}$

Now, to get the water to undergo a phase change, i.e. make it go from solid to liquid at constant temperature, ${0}^{\circ} \text{C}$, you need to supply

${q}_{1} = {m}_{\text{water" * DeltaH_"fus}}$

q_1 = 36color(red)(cancel(color(black)("g"))) * 0.335"kJ"/color(red)(cancel(color(black)("g"))) = "12.06 kJ"

To heat the water from liquid at ${0}^{\circ} \text{C}$ to liquid at ${25}^{\circ} \text{C}$, you need to use the equation

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

${c}_{\text{water}}$ - the specific heat of water;
$\Delta T$ - the change in temperature, calculated at the final temperature minus the initial temperature.

In your case, you have

$\Delta T = 25 - 0 = {25}^{\circ} \text{C}$

This is equal to $\text{25 K}$, since you have

$\Delta T = \left(25 + 273.15\right) - \left(0 + 273.15\right)$

$\Delta T = 25 + \textcolor{red}{\cancel{\textcolor{b l a c k}{273.15}}} - 0 - \textcolor{red}{\cancel{\textcolor{b l a c k}{273.15}}} = \text{25 K}$

Therefore, you get

${q}_{2} = 36 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)("K")))) * (25 - 0)color(red)(cancel(color(black)("K}}}}$

${q}_{2} = \text{3762 J" = "3.762 kJ}$

The total heat needed will be

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

q_"total" = 12.06 + 3.762 = color(green)("15.8 kJ")#

I'll leave the answer rounded to three sig figs, despite the fact that you only gave two sig figs for the mass of the water.