In: Chemistry
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) * |
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