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

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer

can you attach the