Question

At 1 atm, how much energy is required to heat 49.0 g of H2O(s) at –10.0 °C to H2O(g) at 119.0 °C?

Answer 1

Q = mst

Q = energy requried

m = mass of sample (here water)

s = specific heat of water = 4.186 joule/gram °C

t = t₂-t₁ where t₂ is the final temperature and t₁ is the initial temperature

Heat required to raise the temperture from -10 to 0

Q = mst = 49g x 4.186 joule/gram °C x (0-(-10) °C= 2051.14 J

Heat of Fusion of water

H₂O(s)------> H₂O(l ); ΔH_fus = 6.00 kJ mol = 6000 J/mol = 6000J /18g = 333.33 J/g

For 49.0 g the ΔH_fus = 333.33 x 49 = 16333.17 J

Heat required to raise the temperture from 0 to 100°C

Q = mst = 49g x 4.186 joule/gram °C x (100-0) °C= 20511.4 J

Heat of Vaporisation of water

H₂O(l)------> H₂O(v); ΔH_vap = 40.79 kJ mol = 40790J/mol = 40790/18 = 2266.11 J/g

For 49.0 g the ΔH_fus = 2266.11 x 49 = 111039.44 J

Heat required to raise the temperture from 100 to 119 °C

Q = mst = 49g x 4.186 joule/gram °C x (119-100) °C = 3897.16 J

So, Total heat required is =

2051.14 J +16333.17 J+20511.4 J+111039.44 J+ 3897.16 J =153832.31 J = 153.83 kJ

The energy required to heat 49.0 g of H₂O(s) at –10.0 °C to H₂O(g) at 119.0 °C is 153.83 kJ