How much heat (in kj) is evolved in converting 1.00 mol of steam at 130.0 ∘c to ice at -55.0 ∘c? the heat capacity of steam is 2.01 j/g⋅∘c and of ice is 2.09 j/g⋅∘c?

The heat will be evolved in following process:

1) cooling steam from to = Q1

2) conversion of steam to water =Q2

3) Cooling of water to to =Q5

Let us calculate each energy one by one:

1) Q1:

Mass of water vapour = moles X molar mass = 1mol X 18 = 18 grams

heat capacity of steam is

2) Q2:

for this we need heat of vaporisation of water = -40.7kJ/mol

so the heat evolved in conversion of stem to water is -40.7kJ

3) Q3

the specific heat of water is

heat evolved in cooling of water will be

4)Q4:

The heat of conversion of ice to water (melting) is 6.02kJ/mol

5) Q5

The cooling of ice will require

The total energy released is

The heat released will be 59.474 kJ

the formula would be mgso4

mgso[subscript]4

"hydrogen bonding is the attraction of h atoms in the water molecule (slight + charge) to o atoms in a neighboring water molecule (slight - charge); it is a weak interatomic force. but it is still strong enough to the water molecules "stick" together."