Question #319f9

1 Answer
Jun 16, 2017

Answer:

#"600 g"#

Explanation:

Start by looking up the enthalpy of fusion of water

#DeltaH_"fus" = "333.55 J g"^(-1)#

https://en.wikipedia.org/wiki/Enthalpy_of_fusion

Now, the enthalpy of fusion of a given substance tells you the enthalpy change that occurs when #"1 g"# of said substance goes from solid to liquid at its melting point.

In your case, you know that in order to convert #"1 g"# of solid water, i.e. ice, from solid at its normal melting point of #0^@"C"# to liquid water at #0^@"C"#, you need to provide it with #"333.55 J"# of heat.

Now, you know that you are providing an unknown mass of ice with

#200 color(red)(cancel(color(black)("kJ"))) * (10^3color(white)(.)"J")/(1color(red)(cancel(color(black)("kJ")))) = 2 * 10^5# #"J"#

of heat. In order to determine the mass of ice at #0^@"C"# that can be converted to liquid water at #0^@"C"#, i.e. you can use the enthalpy of fusion as a conversion factor.

#2 * 10^5 color(red)(cancel(color(black)("J"))) * overbrace("1 g"/(333.55color(red)(cancel(color(black)("J")))))^(color(blue)("equivalent to" color(white)(.)DeltaH_"fus" = "333.55 J g"^(-1))) = "599.6 g"#

Rounded to one significant figure, the number of sig figs you have for the amount of heat you provide to the sample, the answer will be

#color(darkgreen)(ul(color(black)("mass of ice = 600 g")))#

You can thus say that #"200 kJ"# of heat are enough to convert #"600 g"# of ice at #0^@"C"# to liquid water at #0^@"C"#.