How much heat is released when 105g of steam at 100.0c is cooled to ice at -15c? enthalpy of vaporization of water is 40.67kj/mol, the enthalpy of fusion of water is 6.01kj/mol, the molar heat capacity of water is 75.4j/(mol. c) and the molar heat capacity of ice is 36.4j/(mol. c)

Answer : The heat released is, 319.28 kJ

Solution :

The conversions involved in this process are :

Now we have to calculate the enthalpy change.

where,

= enthalpy change = ?

Mass of water = 105 g

Moles of water =

= specific heat of solid water =

= specific heat of liquid water =

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

= enthalpy change for condensation = 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)

Negative sign indicates that the heat is released during the process.

Therefore, the heat released is, 319.28 KJ

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"

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