Question

Calculate the amount of energy (in kJ) necessary to convert 557 g of liquid water from 0°C to water vapor at 172°C. The molar heat of vaporization (H_vap) of water is 40.79 kJ/mol. The specific heat for water is 4.184 J/g ·°C, and for steam is 1.99 J/g ·°C. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

= KJ

Answer 1

Solution :

We've,

Specific heat of water = 4.184 J/g °C .

Specific heat of steam = 1.99 J/g °C .

Molar heat of vaporization of water = 40.79 kJ/mol .

Amount of energy required to convert 557 grams of water from 0 °C to steam of 172 °C will be sum total of

a) Energy required to raise the temperature of water from 0 °C to 100 °C

b) Energy required to convert water to steam at 100 °C

c) Energy required to raise the temperature of steam from 100 °C to 172 °C

Let's calculate (a) i.e. heat required to raise temperature of water from 0°C to 100 °C

For this, we've a formula -

Q = m × c × ∆T

where, Q = Energy or heat required, m = mass of substance, c = Specific heat & ∆T = change in temperature.

Now,

m = 557 grams.

c = 4.184 J/g °C

∆T = (100 °C - 0 °C) = 100 °C

Therefore, putting values to formula, we get -

Q(a) = m × c × ∆T

= 557 g × 4.184 J/g °C × 100 °C

= ( 557 × 4.184 × 100 ) J

= 233048.8 J or 233.05 kJ. ( as there are 1000 J in 1 kJ)

Q(a) = 233.05 kJ.

Now, we'll calculate for (b), i.e heat required to convert water to steam at 100 °C

For this, we've given molar heat of vaporization of water i.e. equal to 40.79 kJ/mol

This means 40.79 kJ of heat or energy is required to convert 1 mol of water to 1 mol of steam.

OR

For converting 1 mol of water to steam, we need 40.79 kJ. Now, 1 mol of water is equal to 18 grams of water. Hence, we can say for converting 18 grams of water to steam, we need 40.79 kJ

Thus, for converting 1 gram of water to steam we'll need :

of heat or energy

Hence, for converting 557 grams of water to steam we'll need :

= 1262.22 kJ.

Hence, Q(b) = 1262.22 kJ

Now, we'll calculate for part(c), i.e. heat required to raise temperature of steam from 100 °C to 172 °C.

For this we've formula :

Q = m × c × ∆T

where, m = mass of substance(steam) = 557 grams.

c = specific heat (steam) = 1.99 J/g °C

∆T = change in temperature = (172 °C - 100 °C) = 72 °C

Thus, putting values to formula, we get :

Q = m × c × ∆T

= 557 g × 1.99 J / g °C × 72 °C

= (557 × 1.99 × 72) J

= 79806.96 J = 79.81 kJ. (as 1 kJ = 1000 J)

Thus, Q(c) = 79.81 kJ.

Therefore, total energy required to convert water at 0°C to steam at 172 ° C :

Q(a) + Q(b) + Q(c)

= 233.05 kJ + 1262.22 kJ + 79.81 kJ

= (233.05 + 1262.22 + 79.81) kJ

= 1575.08 kJ (Answer)

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