The enthalpy change for converting 1.00 mol of ice at -25.0 ∘c to water at 90.0∘c is kj. 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 h2o, δ hfus = 6.01kj/mol, and δhvap = 40.67 kj/mol.

Answer : The enthalpy change for converting 1 mole of ice at to water at is, 7.712 KJ

Solution :

Process involved in the calculation of enthalpy change :

Now we have to calculate the enthalpy change.

where,

= enthalpy change

m = mass of water =

= specific heat of ice = 2.09 J/gk

= specific heat of water = 4.18 J/gk

= enthalpy change for fusion = 6.01 KJ/mole = 0.00601 J/mole

conversion :

= initial temperature of ice =

= final temperature of ice =

= initial temperature of water =

= final temperature of water =

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

    (1 KJ = 1000 J)

Therefore, the enthalpy change for converting 1 mole of ice at to water at is, 7.712 KJ

wher is the rest of the question

the answer is 0.42 is that is an option. this question provides no info