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

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j

did anyone actually do this? like if anyone has submitted this i rly need .