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
In other words, for equal masses, you need
Similarly, for equal masses, you will give off
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"#