# A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture?

Nov 26, 2015

${11}^{\circ} \text{C}$

#### Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

${q}_{1} + {q}_{2} = - {q}_{3} \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$, where

${q}_{1}$ - the heat absorbed by the solid at ${0}^{\circ} \text{C}$
${q}_{2}$ - the heat absorbed by the liquid at ${0}^{\circ} \text{C}$
${q}_{3}$ - the heat lost by the warmer water sample

The two equations that you will use are

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - heat absorbed/lost
$m$ - the mass of the sample
$c$ - the specific heat of water, equal to 4.18 "J"/("g" ""^@"C")
$\Delta T$ - the change in temperature, defined as final temperature minus initial temperature

and

color(blue)(q = n * DeltaH_"fus")" ", where

$q$ - heat absorbed
$n$ - the number of moles of water
$\Delta {H}_{\text{fus}}$ - the molar heat of fusion of water, equal to $\text{6.01 kJ/mol}$

Use water's molar mass to find how many moles of water you have in the $\text{100.0-g}$ sample

100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"

So, how much heat is needed to allow the sample to go from solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$?

q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"

This means that equation $\textcolor{p u r p \le}{\left(1\right)}$ becomes

$\text{33.36 kJ} + {q}_{2} = - {q}_{3}$

The minus sign for ${q}_{3}$ is used because heat lost carries a negative sign.

So, if ${T}_{\text{f}}$ is the final temperature of the water, you can say that

$\text{33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water}$

More specifically, you have

"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))

"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)

Convert the joules to kilojoules to get

33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)

This is equivalent to

$0.418 \cdot {T}_{\text{f" + 2.717 * T_"f}} = 67.925 - 33.36$

${T}_{\text{f" = 34.565/(0.418 + 2.717) = 11.026^@"C}}$

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

T_"f" = color(green)(11^@"C")