Question

Question #605f9

Answer 1

`"25.0 kJ"`

Explanation:

In order to be able to answer this question, you need to know the value of the enthalpy of fusion for water, which is listed as

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

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

When water freezes at its normal boiling point of `0^@"C"`, heat is being given off to the surroundings. More specifically, for every gram of water that freezes, `"333.55 J"` of heat are being released.

This means that your sample will give off a total of

`75.0 color(red)(cancel(color(black)("g"))) * overbrace("333.55 J"/(1color(red)(cancel(color(black)("g")))))^(color(blue)(= DeltaH_"fus")) = "25016.25 J"`

Rounded to three significant figures and expressed in kilojoules

`color(blue)(ul(color(black)("1 kJ" = 10^3"J")))`

the answer will be

`color(darkgreen)(ul(color(black)("heat released = 25.0 kJ")))`