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
Dec 24, 2016

Answer:

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.