# Question #df236

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to look out for the *phase change* underwent by the ice sample.

More specifically, the ice will **absorb** heat from the hot water to go from *solid* at *liquid* at **final temperature** of the mixture.

All this heat will come from the *hot water* sample, hence the drop in temperature between the initial temperature of

Mathematically, this heat exchange can be expressed like this

#color(blue)(q_"ice" + q_"cold water" = -q_"hot water")#

Here the *minus sign* is used because **heat lost** carries a negative sign.

The two equations that you will use are

#color(blue)(q = m * c * DeltaT)" "# , where

*final temperature* minus *initial temperature*

and

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

*number of moles* of water

*molar heat of fusion* of water

So, your equation will look like this

#n * DeltaH_"fus" + m * c_"water" * DeltaH_"cold" = - m_"water" * c_"water" * DeltaH_"cold"#

Since *moles of water* that were initially present as ice, you can rewrite this term by using water's molar mass and the mass of ice,

#color(blue)(m)color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = color(blue)(m)/18.015" moles H"_2"O"#

The above equation thus becomes

#color(blue)(m)/18.015 * DeltaH_"fus" + color(blue)(m) * 18.015 * c_"water" * DeltaT_"cold" = - m_"water" * c_"water" * DeltaT_"hot"#

Plug in your values and solve for

#color(blue)(m)/18.015 color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("moles"))) + color(blue)(m) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (15 - 0)color(red)(cancel(color(black)(""^@"C"))) = - 64.3color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) * color(red)(cancel(color(black)(""^@"C")))) * (15 - 55)color(red)(cancel(color(black)(""^@"C")))#

**Do not** forget that you're dealing with *kilojoules* **and** *joules*, so convert the kilojoules to joules to keep track of the proper units

#color(blue)(m) * 333.6color(red)(cancel(color(black)("J"))) + color(blue)(m) * 62.7color(red)(cancel(color(black)("J"))) = 10751color(red)(cancel(color(black)("J")))#

This is equivalent to

#color(blue)(m) * 396.3 = 10751 implies color(blue)(m) = 10751/396.3 = "27.128 g"#

I'll leave the value rounded to two sig figs

#m_"ice" = color(green)("27 g")#