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

the answer is 75.0%.

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