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

Answer:

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"#