Question

How much heat does a 23.0 gram ice cube absorb as its temperature increases from-17.4 °C to 0.0 °C ? The specific heat of ice is 2.108 J/(g C).

Answer 1

Here's what I got.

Explanation:

The specific heat of ice tells you how much heat must be absorbed by `"1 g"` of ice in order for its temperature to increase by `1^@"C"`.

You know that since

`c = "2.108 J g"^(-1)"C"^(-1)`

you need to provide `"2.108 J"` of heat per gram of ice in order to increase the temperature of the sample by `1^@"C"`. This means that in order to increase the temperature of `"23.0 g"` of ice by `1^@"C"`, you need to provide

`23.0 color(red)(cancel(color(black)("g"))) * "2.108 J"/(1color(red)(cancel(color(black)("g"))) 1^@"C") = "48.484 J g"^(-1)`

So, you know that for a `"23.0-g"` sample of ice, `"48.484 J"` are needed in order to increase its temperature by `1^@"C"`. In your case, you need the temperature of the sample to increase by

`DeltaT = 0.0^@"C" - (-17.4^@"C") = 17.4^@"C"`

This means that you will need

`17.4 color(red)(cancel(color(black)(""^@"C"))) * "48.484 J"/(1color(red)(cancel(color(black)(""^@"C")))) = color(darkgreen)(ul(color(black)("844 J")))`

I'll leave the answer rounded to three sig figs.

Keep in mind that this much heat is needed in order to get you from ice at `-17.4^@"C"` to ice at `0.0^@"C"`. If you want the melt the ice to liquid at `0.0^@"C"`, you're going to need to provide it with additional heat.