# As a 6 kg of a liquid substance at its freezing point completely freezes , it gives enough heat to melt 3 kg of ice at 0^o C the heat of fusion of the substance is ?

##### 1 Answer
May 15, 2017

Half that of water.

#### Explanation:

The enthalpy of fusion, $\Delta {H}_{\text{fus}}$, tells you the energy needed to convert $\text{1 g}$ of a given substance from solid to liquid at its melting point.

Similarly, you can say that the enthalpy of fusion tells you the energy released when $\text{1 g}$ of a substance goes from liquid to solid at its freezing point.

In your case, you know that when it freezes, a $\text{6-kg}$ sample of an unknown substance give off enough heat to melt $\text{3 kg}$ of ice.

Right from the start, the fact that the two masses are different tells you that it takes different amounts of energy to cause a liquid $\to$ solid or solid $\to$ liquid change of state.

If the unknown substance had the same enthalpy of fusion as water, then the heat given off by the $\text{6-kg}$ sample would melt $\text{6 kg}$ of ice.

$\text{same energy needed + same mass = same"color(white)(.)DeltaH_"fus}$

However, that is not the case here. You need

$\text{6 kg" = color(red)(2) * "3 kg}$

of the unknown substance to get the same energy needed to melt $\text{3 kg}$ of ice, so the enthalpy of fusion will be $\textcolor{red}{2}$ times lower than that of water.

In other words, for equal masses, you need $\textcolor{red}{2}$ times as much heat to melt the ice than to melt the unknown substance.

Similarly, for equal masses, you will give off $\textcolor{red}{2}$ times more heat when the ice freezes than when the unknown substance freezes.

Therefore, you can say that

$\Delta {H}_{\text{fus substance" = 1/color(red)(2) * DeltaH_"fus water}}$

If you want, you can look up the enthalpy of fusion of water and double-check the answer.

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

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

So, $\text{1 g}$ of ice needs $\text{333.55 J}$ of heat to go from solid at ${0}^{\circ} \text{C}$ to liquid at ${0}^{\circ} \text{C}$. This means that the sample will need

3 color(red)(cancel(color(black)("kg"))) * (10^3color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "333.55 J"/(1color(red)(cancel(color(black)("g")))) = 1.00065 * 10^6 $\text{J}$

This much heat is being given off when $\text{6 kg}$ of the unknown substance go from liquid at the freezing point to solid at the freezing point, so the heat given off when $\text{1 g}$ undergoes this phase change will be

1 color(red)(cancel(color(black)("g"))) * (1color(red)(cancel(color(black)("kg"))))/(10^3color(red)(cancel(color(black)("g")))) * (1.00065 * 10^6color(white)("J"))/(6color(red)(cancel(color(black)("kg")))) = "166.775 J"

This means that you have

$\Delta {H}_{\text{fus substance" = "166.775 J g"^(-1) = "333.55 J g"^(-1)/color(red)(2) = 1/color(red)(2) * DeltaH_"fus water}}$