Question

At 1 atm, how much energy is required to heat 63.0 g of H2O(s) at –22.0 °C to H2O(g) at 153.0 °C?

Answer 1

as we know that this process has five steps conversion of H2O, which are

1) H2O (s) at -22⁰ C ------------------> H2O(s) at 0⁰C

E = mst where m = mass; s = specific heat: Δt = temp. difference; E = Heat absorbed or Released

E= 63.0 g x 2.11 J/g/°C x 22 °C

E = 2.9 kJ

2) H2O (s) at 0⁰ C ------------------> H2O(l) at 0⁰C

E= mL where m= mass L = Latenr heat of melting

E= 63g x 334 J/g

= 21042 J

= 21.04kJ.

3) H2O (l) at 0⁰ C ------------------> H2O(l) at 100⁰C

E = mst where m = mass; s = specific heat: Δt = temp. difference; E = Heat absorbed or Released

E= 63g x  4.18 J/g/°C x 100°C

E=26.33 kJ

4)   H2O (l) at 100⁰ C ------------------> H2O(g) at 100⁰C

E= mL where m= mass L = Latenr heat of melting

E= 63g x 2260 J/g

= 142.38 kJ

5) H2O (g) at 100⁰ C ------------------> H2O(g) at 153⁰C

E = mst where m = mass; s = specific heat: Δt = temp. difference; E = Heat absorbed or Released

E= 63g x 2.08 J/g/°C x 53°C

E=6.95 kJ

so total energy required to heat 63.0 g of H2O(s) at –22.0 °C to H2O(g) at 153.0 °C is

E= 2.9 +21.04+26.33+142.38+6.95

E=199.6 kJ