Question

How much energy (kJ) is required to convert a 25.0 g ice cube at -15.°C to water vapor at 250°C?

Answer 1

Ans:

The scenario has to be divided into 5 steps

1. H₂O (s) at -15^oC to H₂O (s) at 0^oC (Thermal energy required, q1)
2. H₂O (s) at 0^oC to H₂O (l) at 0^oC (Thermal energy required, q2)
3. H₂O (l) at 0^oC to H₂O (l) at 100^oC (Thermal energy required, q3)
4. H₂O (l) at 100^oC to H₂O (g) at 100^oC (Thermal energy required, q4)
5. H₂O (g) at 100^oC to H₂O (g) at 250^oC (Thermal energy required, q5)

Total thermal energy required = Q = q1 +q2 + q3 + q4 + q5

Mass of the ice taken, m = 25.0 g

∆T = represents the temperature change in ^oC

∆H_fus= Heat of fusion of water = 0.334 kJ/g (data collected)

∆H_vap = Heat of vaporization at 100^oC = 2.261 kJ/g (data collected)

Specific heat capacity = ‘c’

specific heat capacity of H₂O (g) = 2.01 J g^-1 °C^-1 (data collected)

specific heat capacity of H₂O (l) = 4.18 J g^-1 °C^-1 (data collected)

specific heat capacity of H₂O (s) = 2.22 J g^-1 °C^-1 (data collected)

1	q1 = m. c. ∆T   q1 = 25.0 x 2.22 x (0-(-15)) = 832.5 J
2	q2 = m. ∆Hfus q2 = 25.0 g x 0.334 kJ/g = 8.35 x 103 J
3	q3 = m. c. ∆T q3 = 25.0 x 4.18 x (100-0) = 10450 J
4	q4 = m. ∆Hvap q4 = 25.0 g x 2.261 kJ/g = 56.525 x 103 J
5	q5 = m. c. ∆T   q5 = 25.0 x 2.01 x (250-100) = 7537.5 J
Q = 83695 J = 83.695 kJ

Total energy required for the conversion = 83.695 kJ