# What is the final temperature of the combined system when "10.0 g" of ice is placed into "50.0 g" of water at 32.0^@ "C"? c_"water" = "4.184 J/g"^@ "C", while DeltaH_(fus)^@ = "333 J/g" for ice melting into water.

Oct 22, 2017

$\text{Final temperature} \approx {13.39}^{o} C$

#### Explanation:

According to principle of calorimetry,

$\text{Heat absorbed by ice" + "Heat absorbed by water formed from ice" = "Heat lost by hot water}$

$10 \times 333 + 10 \times 4.18 \times \left({T}_{f} - 0\right) = 50 \times 4.18 \times \left(32 - {T}_{f}\right)$

solving for ${T}_{f}$ ,

${T}_{f} \approx {13.39}^{o} C$

Oct 22, 2017

I got ${13.40}^{\circ} \text{C}$.

Since ice is put into hotter water, it must first melt. The total heat contribution to melt the ice is:

$\textcolor{g r e e n}{{q}_{\text{ice" = m_"ice"DeltabarH_"fus}}}$

where $\Delta {\overline{H}}_{\text{fus}}$ is the mass enthalpy of a phase transition in $\text{J/g}$, and $m$ is the grams of water.

...and its value is:

${q}_{\text{melt" = 10.0 cancel"g ice" xx "333 J"/cancel"g ice" = "3330 J}}$

The contribution to heat it up to some final temperature is going to occur at constant atmospheric pressure but nonconstant temperature, so

$\textcolor{g r e e n}{{q}_{\text{ice" = overbrace(m_"ice")^("as melted water")c_"water"overbrace(DeltaT_"ice")^"as melted water}}}$

We are saying:

${\overbrace{\text{ice")^(0.00^@ "C") stackrel("Based on "DeltaH_"fus"" ")(->) overbrace("water")^(0.00^@ "C") stackrel("Based on specific heat capacity"" ")(->) overbrace("water}}}^{{T}_{f}}$

And the contribution to cooling the hotter water has a magnitude of:

$\textcolor{g r e e n}{{q}_{\text{water" = m_"water"c_"water"DeltaT_"water}}}$

By conservation of energy, the thermal energy from the hotter water goes into melting the ice and into heating it. So:

${q}_{\text{melt" + q_"ice" + q_"water}} = 0$

$\implies {q}_{\text{melt" + q_"ice" = -q_"water}}$

$\implies {m}_{\text{ice"DeltabarH_"fus" + overbrace(m_"ice")^("as melted water")c_"water"overbrace(DeltaT_"ice")^"as melted water}}$

$= - {m}_{\text{water"c_"water"DeltaT_"water}}$

Expanding this out, we then get:

"3330 J" + "10.0 g ice" cdot "4.184 J/g"^@ "C" cdot (T_f - 0.00^@ "C")

= -"50.0 g water" cdot "4.184 J/g"^@ "C" cdot (T_f - 32.0^@ "C")

We know the units will work out since this is all in $\text{J}$, $\text{g}$, and $\text{^@ "C}$, and it turns out that distributing terms is easier than working this out algebraically with no numbers.

We now omit the units on purpose and we'll put them back later.

$3330 + 41.84 {T}_{f} = - 209.2 \left({T}_{f} - 32\right)$

Distribute terms to get:

$3330 + 41.84 {T}_{f} + 209.2 {T}_{f} = 6694.4$

Solve for ${T}_{f}$, the equilibrium temperature (which MUST be between ${0.00}^{\circ} \text{C}$ and ${32.00}^{\circ} \text{C}$; why?):

$3330 + 251.04 {T}_{f} = 6694.4$

$\implies 251.04 {T}_{f} = 3364.4$

$\implies \textcolor{b l u e}{{T}_{f}} = \left(3364.4 \cancel{\text{J/g")/(251.04 cancel"J/g"cdot""^@ "C}}\right)$

$= \underline{\textcolor{b l u e}{{13.40}^{\circ} \text{C}}}$