Calculate the enthalpy change associated with the conversion of 25.0 grams of ice at -4.00 °c to water vapor at 110.0 °c. the specific heats of ice, water, and steam are 2.09 j/g-k, 4.18 j/g-k, and 1.84 j/g-k, respectively. for , δhfus = 6.01 kj/mol and δhvap = 40.67 kj/mol.

Answer : The enthalpy change is, 104.5327 KJ

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change = ?

m = mass of water = 25 g

= specific heat of solid water = 2.09 J/gk

= specific heat of liquid water = 4.18 J/gk

= specific heat of liquid water = 1.84 J/gk

n = number of moles of water =

= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole

= enthalpy change for vaporization = 40.67 KJ/mole = 40670 J/mole

Now put all the given values in the above expression, we get

    (1 KJ = 1000 J)

Therefore, the enthalpy change is, 104.5327 KJ

hydrocarbons are organic compounds that contain only hydrogens and carbons. they can be saturated (those which contain only carbon-carbon single bonds) or unsaturated (those which contain carbon-carbon double or triple bonds) in nature. hydrocarbons can undergo several reactions like substitution, elimination etc.  

on substituting one or more hydrogen atoms in hydrocarbon it results in the formation of haloalkane. as the halogen atoms are large compared to the carbon and hydrogen atoms, due to which the molecular weight increases and the bond become polar because of presence electronegative halogen atom and results in the increase in boiling point of the haloalkane.

thus, the correct option is (b) that is the boiling point of the new compound increases on substituting a hydrogen atom with a halogen in a hydrocarbon.