Question

In: Chemistry

At 1 atm, how much energy is required to heat 37.0 g of H2O(s) at –10.0...

At 1 atm, how much energy is required to heat 37.0 g of H2O(s) at –10.0 °C to H2O(g) at 169.0 °C? Helpful constants can be found here.

Quantity per gram per mole
Enthalpy of fusion 333.6 J/g 6010. J/mol
Enthalpy of vaporization 2257 J/g 40660 J/mol
Specific heat of solid H2O (ice) 2.087 J/(g·°C) * 37.60 J/(mol·°C) *
Specific heat of liquid H2O (water) 4.184 J/(g·°C) * 75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam) 2.000 J/(g·°C) * 36.03 J/(mol·°C) *

Solutions

Expert Solution

This is calculated in 5 stages.

1) To heat the ice from -10°C to 0°C (dT = 10°C)

Q = mcdT

         Q = heat ,

m = mass of water = 37 g

        c = specific heat of ice = 2.087 J/g/°C

        dT = temperature difference = 10°C

Q = 37 g x 2.087 J/g/°C x 10°C = 772.2 J

Q = 772.2 J

2) To melt the ice at 0°C to water at 0°C.

Q = m x Enthalpy of fusion

    = 37 g x 333.6 J/g

    = 12343.2 J

Q = 12343.2 J

3) To heat the water from 0°C to boiling at 100°C

Q = mcdT where c = specific heat of water , dT = 100°C

   = 37 g x 4.184 J/g/°C x 100°C

= 15480.8 J

Q = 15480.8 J

4) To vaporize the water to steam at 100°C
Q = m x Enthalpy of vaporization

    = 37 g x 2257 J/g

    = 83509 J

Q = 83509 J

5) To heat the steam from 100°C to 169°C

Q = mcdT where c = specific heat of gaseous H2O (steam) , dT = 69°C

   = 37 g x 2.000 J/g/°C x 69 °C

= 5106 J

Q = 5106 J

Therefore,

Total heat energy required = 772.2 J + 12343.2 J + 15480.8 J + 83509 J + 5106 J

                                     = 117211.2 J

                                    = 117.2 kJ


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