Some water and 270g of ice is given the same amount of heat. What amount of water can be vapoured by the time the ice melts fully?

this is from method of mixtures

Jun 22, 2018

You can vaporize 39.9 g of water.

Explanation:

$\text{Heat to melt ice = Heat to vaporize water}$

$\textcolor{w h i t e}{m m m m} {q}_{1} \textcolor{w h i t e}{m m l l} = \textcolor{w h i t e}{m m m m m} {q}_{2}$

color(white)(mml)m_1Δ_text(fus)H color(white)(m)= color(white)(mmm)m_2Δ_text(vap)H

where

"m_1color(white)(mm) = the mass of the ice

Δ_text(fus)Hcolor(white)(l) =  the enthalpy of fusion of ice

${m}_{2} \textcolor{w h i t e}{m m} =$ the mass of the water

Δ_text(vap)H =  the enthalpy of vaporization of water

In this problem,

${m}_{1} \textcolor{w h i t e}{m m} = \text{270 g}$

Δ_text(fus)Hcolor(white)(l) = "333.55 J·g"^"-1"

m_2color(white)(mm) = ?

Δ_text(vap)H = "2256.3 J·g"^"-1"

m_1Δ_text(fus)H = m_2Δ_text(vap)H

"270 g" × 333.55 color(red)(cancel(color(black)("J·g"^"-1"))) = m_2 × 2256.3 color(red)(cancel(color(black)("J·g"^"-1")))

${m}_{2} = \text{90 060 g"/2256.3 = "39.9 g}$

Jun 22, 2018

Here is an alternative approach. I get about $\text{40.0 g}$.

From your notes, you should have that the enthalpies of fusion and vaporization for water at the freezing point and boiling point respectively are:

$\Delta {H}_{f u s} = \text{6.02 kJ/mol}$
$\Delta {H}_{v a p} = \text{40.67 kJ/mol}$

Since

• the two masses of ice and water are given the same amount of heat $q$ in $\text{kJ}$,
• water has the same molar mass as ice,
• more heat is needed to vaporize the same mass of water than is needed to melt the same mass of ice

...we expect that the mass of water that can be vaporized, if enough heat is used to melt a certain mass of ice, would be a smaller mass.

From the units, we find:

overbrace("mass of ice" xx "mol water"/"g water" xx "6.02 kJ"/"mol ice")^"heat to melt ice" = overbrace("mass of water" xx "mol water"/"g water" xx "40.67 kJ"/"mol water")^"heat to vaporize water"

or in terms of variables,

overbrace(m_"ice"/cancel(M_"water")DeltaH_(fus))^"heat to melt ice" = overbrace(m_"water"/cancel(M_"water")DeltaH_(vap))^"heat to vaporize water"

The amount of ice that can be melted divided by the amount of water that can be vaporized is therefore given by the ratio of the enthalpies:

(DeltaH_(vap))/(DeltaH_(fus)) = m_"ice"/m_"water" = ("mass of ice melted")/("mass of water vaporized")

$\implies \left(\text{40.67 kJ/mol")/("6.02 kJ/mol") = "270.0 g"/("mass of water vaporized}\right)$

Thus, if $270.0$ $g$ of ice would have been melted, then about $40.0$ $g$ of water can be vaporized.