# 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

Half that of water.

#### Explanation:

The **enthalpy of fusion**,

Similarly, you can say that the enthalpy of fusion tells you the energy released when

In your case, you know that when it freezes, a

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

If the unknown substance had the same enthalpy of fusion as water, then the heat given off by the

#"same energy needed + same mass = same"color(white)(.)DeltaH_"fus"#

However, that is not the case here. You need

#"6 kg" = color(red)(2) * "3 kg"#

of the unknown substance to get the same energy needed to melt **times lower** than that of water.

In other words, for **equal masses**, you need **times as much heat** to melt the ice than to melt the unknown substance.

Similarly, for **equal masses**, you will give off **times more heat** when the ice freezes than when the unknown substance freezes.

Therefore, you can say that

#DeltaH_"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.

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

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

So,

#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# #"J"#

This much heat is being given off when

#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

#DeltaH_"fus substance" = "166.775 J g"^(-1) = "333.55 J g"^(-1)/color(red)(2) = 1/color(red)(2) * DeltaH_"fus water"#