Question

Question #35b3b

Answer 1

`6.0 * 10^2` `"J"`

Explanation:

The key here is the enthalpy of fusion for ice, which is given to you as

`DeltaH_"fus" = color(blue)("333.55 J") color(white)(.)"g"^(-1)`

For a given substance, the enthalpy of fusion tells you the amount of heat needed in order to convert `"1 g"` of the substance from solid at its melting point to liquid at its melting point.

In your case, you know that in order to turn `"1 g"` of ice at `0^@"C"`, the normal melting point of ice, to liquid water at `0^@"C"`, you need `color(blue)("333.55 J")` of heat.

This means that your sample will require

`1.8 color(red)(cancel(color(black)("g"))) * color(blue)("333.55 J")/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)(6.0 * 10^2color(white)(.)"J")))`

The answer is rounded to two sig figs, the number of sig figs you have for the mass of the sample.