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

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the maximum mass of nh3 is nh3 in grams