Question

Please Help Me - Clausius-Clapeyron Equations?

How much heat (in kJ) is required to convert 423 g of liquid H2O at 24.0°C into steam at 152°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C . The heat of vaporization (ΔHvap) is 40.65 kJ/mol.)

Answer 1

The total energy required is 1135.5 kJ

Explanation:

There are three separate events occurring here, and each must have its energy calculated separately:

1. Water is warmed from 24.0 °C to 100 °C:

   This calculation is:
   `q=mC_sDeltaT = 423g xx4.184J/(g °C) xx 76 °C= 134507 J=134.5 kJ`

2. Next, the water must be converted to steam at 100 °C. Since we have the molar heat of vapourization, we need to note that 423 g is 23.5 moles (`423-:18`)

   `q=23.5 "mol" xx 40.65 (kJ)/"mol" = 955.3 kJ`

3. Finally, the steam is heated from 100 °C to 152 °C

   `q= 423 xx 2.078 J/(g °C) xx 52 °C = 45708 J = 45.7 kJ`

So, all together, the energy required is `134.5+955.3+45.7 = 1135.5kJ`

Answer 2

Heat Required`= 1135.5kJ`

Explanation:

Calm down.. take these problems ONE step at a time, and they become simple arithmetic.

FIRST - warm the water. We have liquid water, so the first heat required is to increase it from `24.0^oC` to boiling at `100.0^oC`. For this we will use the heat capacity of liquid water, `4.184 J/(g·°C)`. Make sure the units match up, and calculate.

`423g xx (100 - 24) xx 4.184 J/(g·°C) = 134507J`

Or, `134.5kJ`

SECOND - vaporize the water. For this we use the heat of vaporization: `40.65 "kJ"/"mol"`.

`423g xx "1 mol"/("18 g H"_2"O") xx 40.65 = 955.3kJ`

THIRD - heat the steam using the heat capacity of steam, 2.078 J/g·°C,
`423g xx (152 - 100) xx 2.078 J/(g·°C) = 45708J`

Or, `45.7kJ`

FINALLY, add them all up!

Heat Required = `134.5 + 955.3 + 45.7 = 1135.5kJ`