Question

Question #5141e

Answer 1

`"14,800 J"`

Explanation:

The idea here is that you're going to need enough heat to

1. Heat `"4.80 g"` of ice at `-30.0^@"C"` to ice at `0^@"C"`
2. Convert the sample from ice at `0^@"C"` to liquid water at `0^@"C"`, i.e. perform a solid `->` liquid phase change
3. Heat `"4.80 g"` of liquid water at `0^@"C"` to liquid water at `100^@"C"`
4. Convert the sample from liquid water at `100^@"C"` to vapor at `100^@"C"`, i.e. perform a liquid `->` vapor phase change

The total heat needed to get the sample from ice at `-30.0^@"C"` to vapor at `100^@"C"` will be the sum of the heat needed to ensure that every step in this series takes place.

`q_"total" = sumq_"each step"`

Now, for each step, you will need to use

1. The specific heat of ice
2. The heat of fusion of water
3. The specific heat of water
4. The heat of vaporization of water

The two equations that you will use are

`color(blue)(ul(color(black)(q = m * DeltaH_"fus/vap")))" " " "color(darkorange)((1))`

This equation is used for phase changes, i.e. when the change in temperature is equal to zero. Here `m` is the mass of the sample and `DeltaH_"fus/vap"` is the heat of fusion.vaporization, depending on which phase change you're dealing with.

`color(blue)(ul(color(black)(q = m * c_"ice/water" * DeltaT)))" " " "color(darkorange)((2))`

This equation is used when you're heating the sample without performing a phase change. Here `c` is the specific heat of ice/water and `DeltaT` is the change in temperature

This means that, for each step, you will use

1. Use equation `color(darkorange)((2))` with `c_"ice"` and a change in temperature of `DeltaT = 0^@"C" - (-30.0^@"C") = 30.0^@"C"`
2. Use equation `color(darkorange)((1))` with `DeltaH_"fus"`
3. Use equation `color(darkorange)((2))` with `c_"water"` and a change in temperature of `DeltaT = 100^@"C" - 0^@"C" = 100^@"C"`
4. Use equation `color(darkorange)((1))` with `DeltaH_"vap"`

So, you can put all this together to get

`q_"total" = overbrace(4.80 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 30.0color(red)(cancel(color(black)(""^@"C"))))^(color(blue)("step 1")) + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "334.16 J" color(red)(cancel(color(black)("g"^(-1)))))^(color(blue)("step 2"))`

`color(white)(q_"total") + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 100.0color(red)(cancel(color(black)(""^@"C"))))^(color(blue)("step 3")) + overbrace(4.80 color(red)(cancel(color(black)("g"))) * "2259J" color(red)(cancel(color(black)("g"^(-1)))))^(color(blue)("step 4"))`

`color(darkgreen)(ul(color(black)(q_"total" = "14,800 J")))`

The answer is rounded to three sig figs.