posted by Pete on .
Topic: Sublimation and Fusion
How much heat (in kJ) is required to warm 13.0 g of ice, initially at -14.0 C, to steam at 112.0 °C? The heat capacity of ice is 2.09 J/g⋅°C and that of steam is 2.01 J/g⋅°C.
Any help on how to start this is really appreciated.
Do this in several steps.
heat to raise T ice at -14 C to zero.
heat to melt ice at zero.
heat to raise T from zero to 100 C.
heat to vaporize water at 100.
heat to raise steam from 100 to 112 C.
The total is the sum of each part above.
Post your work if you get stuck.
Step 1: Heat to raise T of ice at -14.0 °C to 0.0 °C
q_1 = m⋅C_s,ice⋅ΔT
q_1 = (13.0 g)(2.09 J/g⋅°C)[0.0 °C - (-14.0 °C)]
q_1 = 380.38 J
q_1 = 0.38038 kJ
Step 2: Heat to melt ice at 0.0 °C
q_2 = n⋅ΔH_fus
q_2 = [13.0 g⋅(1 mol/18.016 g)⋅(6.02 kJ/mol)
q_2 = [13.0 g⋅(1 mol/18.016 g)]⋅(6.02 kJ/mol)
q_2 = 4.343916519 kJ
Step 3: Heat to raise T of water from 0.0 °C to 100.0 °C
q_3 = m⋅C_s,liq⋅ΔT
q_3 = (13.0 g)(4.18 J/g⋅°C)(100.0 °C - 0.0 °C)
q_3 = 5434 J
q_3 = 5.434 kJ
Step 4: Heat to vaporize water at 100.0 °C
q_4 = n⋅ΔH_vap
q_4 = [13.0 g⋅(1 mol/18.016 g)⋅(40.7 kJ/mol)
q_4 = [13.0 g⋅(1 mol/18.016 g)]⋅(40.7 kJ/mol)
q_4 = 29.36833925 kJ
Step 5: Heat to raise T of water from 0.0 °C to 100.0 °C
q_5 = m⋅C_s,steam⋅ΔT
q_5 = (13.0 g)(2.01 J/g⋅°C)(112.0 °C - 100.0 °C)
q_5 = 313.56 J
q_5 = 0.31356 kJ
Total Heat Required
q_1 + q_2 + q_3 + q_4 + q_5
(0.38038 kJ) + (4.343916519 kJ) + (5.434 kJ) + (29.3683392 kJ) + (0.31356 kJ)
I'm fairly certain I used the correct values for the heat capacities and the heat of fusion and heat of vaporization. Is this correct?
The procedure looks ok. I didn't look up the Cp values but I know 4.18 is ok for liquid water. I didn't check the math but I note that you need to watch the number of significant figures. I think 3 s.f. are allowable.