# Question 605f9

Feb 21, 2017

$\text{25.0 kJ}$

#### 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

$\Delta {H}_{\text{fus" = "333.55 J g}}^{- 1}$

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

When water freezes at its normal boiling point of ${0}^{\circ} \text{C}$, heat is being given off to the surroundings. More specifically, for every gram of water that freezes, $\text{333.55 J}$ 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

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{1 kJ" = 10^3"J}}}}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{heat released = 25.0 kJ}}}}$