Question

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?

Answer 1

The total heat evolved in converting 1.00 mol of steam at 130.0°C to ice at -55.0°C is 10.70 kJ.

Step-by-step explanation:

To calculate the heat evolved in converting 1.00 mol of steam at 130.0°C to ice at -55.0°C, we need to consider two processes: the cooling of steam to 0°C and the freezing of water at 0°C. Let's break down the calculation step by step.

Cooling of steam to 0°C:

The heat absorbed during this process can be calculated using the formula:

Q = mcΔT

where

Q is the heat absorbed (in J)

m is the mass of the substance (in g)

c is the specific heat capacity (in J/g·°C)

ΔT is the change in temperature (in °C).

We are given that the heat capacity of steam is 2.01 J/g·°C and we have 1.00 mol of steam.

The molar mass of steam is approximately 18.02 g/mol, so the mass of 1.00 mol of steam is 18.02 g.

The change in temperature is 130.0°C - 0°C = 130.0°C.

Using the formula:

Q = (18.02 g)(2.01 J/g·°C)(130.0°C - 0°C)

Q = 4691.46 J = 4.69 kJ

Freezing of water at 0°C:

The heat evolved during this process is the molar heat of fusion, which is the amount of heat required to convert 1 mol of a substance from solid to liquid or vice versa

The molar heat of fusion of water is 6.01 kJ/mol. Since we have 1.00 mol of water, the heat evolved is 6.01 kJ.

Now, we can calculate the total heat evolved:

Total heat evolved = Heat absorbed during cooling + Heat evolved during freezing

Total heat evolved = 4.69 kJ + 6.01 kJ

Total heat evolved = 10.70 kJ

So therefore the total heat evolved is 10.70 kJ

Answer 2

The heat will be evolved in following process:

1) cooling steam from
`130^(0)C` to
`100^(0)C` = Q1

2) conversion of steam to water =Q2

3) Cooling of water to
`0^(0)C[/tex=Q3]</p><p>4) conversion of water to ice=Q4</p><p>5) cooling of ice from [tex]0^(0)C` to
`-55^(0)C`=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
`2.01(J)/(g^(0)C )`

`Q_(1)=massXheat capacityXchangeintemperature=18X2.01X(130-100)=-1085.4J`

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
`4.18(J)/(g^(0)C )`

heat evolved in cooling of water will be
`Q_(3)=massXspecific.heatXchange.in.temperature=18X4.184X100=7531.2J=7.531kJ`

4)Q4:

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

5) Q5

The cooling of ice will require

`Q_(5)=massXspecificheatXchangeintemperature=18X2.09X(-55)=-2069.1J=-2.069kJ`

The total energy released is

`Q_(total)=Q1+Q2+Q3+Q4+Q5=1.085+40.7+7.531+6.02+2.069=<strong>59.474kJ</strong>`

The heat released will be 59.474 kJ