# Question #fc739

Jun 29, 2017

$2.00 \cdot {10}^{3}$ $\text{kJ}$

#### Explanation:

In order to be able to answer this question, you must know the value of the enthalpy of fusion of water

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

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

Now, the enthalpy of fusion tells you the amount of heat needed in order to melt $\text{1 g}$ of a given substance at its melting point.

In your case, you know that in order to convert $\text{1 g}$ of solid water (ice) at ${0}^{\circ} \text{C}$ to $\text{1 g}$ of liquid water at ${0}^{\circ} \text{C}$, i.e. to melt $\text{1 g}$ of ice at its normal melting point, you must provide $\text{333.55 J}$ of heat.

Since you know that

$\text{1 kJ" = 10^3color(white)(.)"J" " }$ and $\text{ " "1 kg" = 10^3color(white)(.)"g}$

you can convert the enthalpy of fusion from joules per gram to kilojoules per kilogram

$\Delta {H}_{\text{vap" = (333.55 color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("J")))) * (color(blue)(cancel(color(black)(10^3))) color(red)(cancel(color(black)("J"))))/"1 kJ" * "1 kg"/(color(blue)(cancel(color(black)(10^3)))color(red)(cancel(color(black)("g")))) = "333.55 kJ kg}}^{- 1}$

So, you know that you need $\text{333.55 kJ}$ of heat in order to melt $\text{1 kg}$ of ice at its normal melting point.

This means that your sample will require

$6.00 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{kg"))) * "333.55 kJ"/(1color(red)(cancel(color(black)("kg")))) = color(darkgreen)(ul(color(black)(2.00 * 10^(3)color(white)(.)"kJ}}}}$

The answer is rounded to three sig figs, the number of sig figs you have for your sample.

You can thus say that the change in enthalpy that accompanies the conversion of $\text{6 kg}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$ is equal to

$\Delta {H}_{\text{fus for 6 kg ice" = +2.00 * 10^3color(white)(.)"kJ}}$

The value must be positive because the heat is being absorbed by the ice.