Question

Question #8e7c8

Answer 1

`"39,800 J"`

Explanation:

The idea here is that you need to calculate how much heat is required in order to

• convert your sample of ice at `-35^@"C"` to ice at `0^@"C"`
• convert the ice at `0^@"C"` to liquid water at `0^@"C"`

In order to be able to calculate the heat required in both cases, you need to know the specific heat of ice and the heat of fusion of water

`c_"ice" = "2.06 J g"^(-1)""^@"C"^(-1)`

`DeltaH_"fus" = "334.16 J g"^(-1)`

So, a substance's specific heat tells you how much heat is needed in order to increase the temperature of `"1 g"` of that substance by `1^@"C"`.

In this case, ice has a specific heat of `"2.06 J g"^(-1)""^@"C"^(-1)`, which means that in order to increase the temperature of `"1 g"` of ice by `1^@"C"`, you need to provide it with `"2.06 J"` of heat.

Your tool of choice here will be this equation

`color(blue)(|bar(ul(color(white)(a/a)q = m * c * DeltaTcolor(white)(a/a)|)))" "`, where

`q` - the amount of heat gained
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as the difference between the final temperature and the initial temperature

In your case, you have a change in temperature of

`DeltaT = 0^@"C" - (-35^@"C") = 35^@"C"`

Plug in your values to find

`q_1 = 98.0 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * 35color(red)(cancel(color(black)(""^@"C")))`

`q_1 = "7,065.8 J"`

Now, adding this much heat to `"98.0 g"` of ice at `-35^@"C"` will get you ice at `0^@"C"`. In order to melt the ice, you must provide it with enough heat to allow it to undergo a solid `->` liquid phase change.

The heat of fusions, `DeltaH_"fus"`, tells you how much heat must be added per gram of solid ice at `0^@"C"` in order to convert it to liquid water at `0^@"C"`.

Since you have `"98.0 g"` of ice at `0^@"C"`, you will need

`98.0 color(red)(cancel(color(black)("g"))) * overbrace("334.16 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)(=DeltaH_"fus")) = "32,748 J"`

The total amount of heat needed to get your sample from ice at `-35^@"C"` to liquid water at `0^@"C"` will thus be

`q_"total" = "7,065.8 J" + "32,748 J"`

`q_"total" = "39,814 J"`

I'll leave the answer rounded to three sig figs

`"the amount of heat needed" = color(green)(|bar(ul(color(white)(a/a)color(black)("39,800 J")color(white)(a/a)|)))`