Question #35b3b

1 Answer
Sep 21, 2017

Answer:

#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.