Question

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 ?

Answer 1

Half that of water.

Explanation:

The enthalpy of fusion, `DeltaH_"fus"`, tells you the energy needed to convert `"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 `"1 g"` of a substance goes from liquid to solid at its freezing point.

In your case, you know that when it freezes, a `"6-kg"` sample of an unknown substance give off enough heat to melt `"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 `->` solid or solid `->` liquid change of state.

If the unknown substance had the same enthalpy of fusion as water, then the heat given off by the `"6-kg"` sample would melt `"6 kg"` of ice.

`"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 `"3 kg"` of ice, so the enthalpy of fusion will be `color(red)(2)` times lower than that of water.

In other words, for equal masses, you need `color(red)(2)` times as much heat to melt the ice than to melt the unknown substance.

Similarly, for equal masses, you will give off `color(red)(2)` 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, `"1 g"` of ice needs `"333.55 J"` of heat to go from solid at `0^@"C"` to liquid at `0^@"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` `"J"`

This much heat is being given off when `"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 `"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

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