Question

Question #df236

Answer 1

`"27 g"`

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 `0^@"C"` to liquid at `0^@"C"`, then the liquid water at `0^@"C"` will absorb heat to go to liquid at `15^@"C"`, the 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 `55^@"C"` and the final temperature of `15^@"C"`.

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

`q` - heat absorbed
`m` - the mass of the sample
`c` - the specific heat of water
`DeltaT` - 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
`DeltaH_"fus"` - the 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 `n` represents 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(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)`, the mass of ice

`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")`