Question

A 100.0g ice cube at 0.0°C is placed in 650g of water at 25°C . what is the final temperature of the mixture?

Answer 1

`11^@"C"`

Explanation:

As far as solving this problem goes, it is very important that you do not forget to account for the phase change underwent by the solid water at `0^@"C"` to liquid at `0^@"C"`.

The heat needed to melt the solid at its melting point will come from the warmer water sample. This means that you have

`q_1 + q_2 = - q_3" " " "color(purple)((1))`, where

`q_1` - the heat absorbed by the solid at `0^@"C"`
`q_2` - the heat absorbed by the liquid at `0^@"C"`
`q_3` - the heat lost by the warmer water sample

The two equations that you will use are

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

`q` - heat absorbed/lost
`m` - the mass of the sample
`c` - the specific heat of water, equal to `4.18 "J"/("g" ""^@"C")`
`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, equal to `"6.01 kJ/mol"`

Use water's molar mass to find how many moles of water you have in the `"100.0-g"` sample

`100.0color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "5.551 moles H"_2"O"`

So, how much heat is needed to allow the sample to go from solid at `0^@"C"` to liquid at `0^@"C"`?

`q_1 = 5.551color(red)(cancel(color(black)("moles"))) * 6.01"kJ"/color(red)(cancel(color(black)("mole"))) = "33.36 kJ"`

This means that equation `color(purple)((1))` becomes

`"33.36 kJ" + q_2 = -q_3`

The minus sign for `q_3` is used because heat lost carries a negative sign.

So, if `T_"f"` is the final temperature of the water, you can say that

`"33.36 kJ" + m_"sample" * c * DeltaT_"sample" = - m_"water" * c * DeltaT_"water"`

More specifically, you have

`"33.36 kJ" + 100.0color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 0)color(red)(cancel(color(black)(""^@"C"))) = -650color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (T_"f" - 25)color(red)(cancel(color(black)(""^@"C")))`

`"33.36 kJ" + "418 J" * (T_"f" - 0) = - "2717 J" * (T_"f" - 25)`

Convert the joules to kilojoules to get

`33.36 color(red)(cancel(color(black)("kJ"))) + 0.418 color(red)(cancel(color(black)("kJ"))) * T_"f" = -2.717 color(red)(cancel(color(black)("kJ"))) * (T_"f" - 25)`

This is equivalent to

`0.418 * T_"f" + 2.717 * T_"f" = 67.925 - 33.36`

`T_"f" = 34.565/(0.418 + 2.717) = 11.026^@"C"`

Rounded to two sig figs, the number of sig figs you have for the mass of warmer water, the answer will be

`T_"f" = color(green)(11^@"C")`