How much heat (in kJ) is required to warm 10.0 g of ice from 10 below freezing to 10 above boiling?

Question

How much heat (in kJ) is required to warm 10.0 g of ice, initially at -10.0°C, to steam at 110°C? Heat capacity of ice is 2.09 J/g•°C, heat capacity of steam is 2.01 J/g•°C.

Answer 1

This is a 5 step problem, where you have to find the energy absorbed at each and sum them all together.

The (5) steps are: 1) ice raised to melting point of 0°C ( q = mcΔT, where c = 2.09 J/g•°C)

2) melting of ice into liquid water (ΔH_fus = 6.02 kJ/ mol)

3) water raised to boiling point at 100°C ( q = mcΔT, where c = 4.184 J/g•C)

4) phase change of water to steam *Boiling!* (ΔH_vap = 40.7 kJ/ mol)

5) steam raised to temperature 110°C ( q = mcΔT, where c is 2.01 J/g•°C)

10.0 g H2O x (1 mol H2O/ 18 g H2O) = 0.5556 mol H2O.

1) q_ice = (10.0 g)(2.09 J/g•°C)(10°C) = 209 J

2) ΔH_fus = 6.02 kJ/mol x 0.5556 mol H2O = 3.344 kJ

3) q_water = (10.0 g)(4.184 J/g•°C)(100°C) = 4184 J

4) ΔH_vap = 40.7 kJ/mol x 0.5556 mol H2O = 22.61 kJ

5) q_steam = (10.0 g)(2.01 J/g•°C)(10°C) = 201 J

Finally summing 1-5 (everything converted into Joules):

209 J + 3.344_E3 J + 4184 J + 22.61_E3 J + 201 J = 30548 J = 30.548 kJ = 30.5 kJ (underlines above for last sig fig)