Question

Question #067de

Answer 1

`"730,000 J"`

Explanation:

All you have to do here is to use the density of liquid oxygen given to you to figure out the mass of the sample.

You know that liquid oxygen at its boiling point has a density of `"1.14 kg L"^(-1)`, which means that `"1 L"` of liquid oxygen at its boiling point has a mass of `"1.14 kg"`.

This means that your sample will have a mass of

`3.0 color(red)(cancel(color(black)("L"))) * "1.14 kg"/(1color(red)(cancel(color(black)("L")))) = "3.42 kg"`

Now, the heat of vaporization tells you the amount of heat needed in order to boil `1` unit of mass of a given substance at its boiling point.

In your case, liquid oxygen is said to have a heat of vaporization equal to `"213 kJ kg"^(-1)`, which means that in order to convert `"1 kg"` of liquid oxygen at its boiling point to oxygen gas at its boiling point, you need to supply `"213 kJ"` of heat.

You can thus say that your sample will require

`3.42 color(red)(cancel(color(black)("kg"))) * "213 kJ"/(1color(red)(cancel(color(black)("kg")))) = "728.5 kJ"`

Rounded to two sig figs, the number of sig figs you have for the volume of the sample, and expressed in joules, the answer will be--keep in mind that you have `"1 kJ" = 10^3` `"J"`

`728.5 color(red)(cancel(color(black)("kJ"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) = color(darkgreen)(ul(color(black)("730,000 J")))`