# 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).

##### 1 Answer

Here's what I got.

#### Explanation:

The **specific heat** of ice tells you how much heat must be absorbed by

You know that since

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

you need to provide **per gram** of ice in order to increase the temperature of the sample by

#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

#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 **ice** at *melt* the ice to **liquid** at