# What minimum mass of ice is needed so that a glass of water at 25^@ "C" is cooled down to 0^@ "C" before the ice completely melts?

Dec 3, 2015

Well, we know what the change in temperature of the water should be, and we want to figure out how much ice should be added to accomplish that without completely melting the ice.

Therefore, we need the amount of energy taken out of the water during that temperature change but not including any phase changes of the water.

${q}_{w} = {m}_{w} {c}_{w} \left({T}_{f} - {T}_{i}\right)$

$= \left(\text{160." cancel"g")(4.18 xx 10^(-3) "kJ/"cancel"g"^@ cancel("C"))(0^@ cancel"C" - 25^@ cancel"C}\right)$

$= - \text{16.720 kJ}$

The sign makes sense since when heat is released, something gets colder.

(The intervals in the $\text{K}$ scale are the same as those in the $\text{^@ "C}$ scale)

Now, all we need to figure out is, "What specific amount of ice absorbing a specific amount of heat $q$ at a constant pressure (${q}_{p} = \Delta {H}_{\text{fus}}$) can absorb that much heat without melting all the way?" The latent heat equation is:

q_"ice" = color(green)(q_w = overbrace(m_"ice")^("?")*DeltaH_"fus,ice")

Heat is transferred from the water to the ice.

"16.720 kJ" = m_"ice"*("6.01 kJ"/"mol" xx ("1 mol H"_2"O")/("18.015 g H"_2"O"))

("16.720" cancel"kJ")/((6.01 cancel"kJ")/cancel"mol" xx (1 cancel"mol H"_2"O")/("18.015 g H"_2"O")) = m_"ice"

$\textcolor{b l u e}{{m}_{\text{ice" ~~ "50.1 g}}}$