Question

Question #319f9

Answer 1

`"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"`.