Consider heating solid water (ice) until it becomes liquid and then gas (steam).
ID: 515721 • Letter: C
Question
Consider heating solid water (ice) until it becomes liquid and then gas (steam). Alternatively, consider the reverse process, cooling steam until it becomes water and, finally, ice. In each case, two types of transitions occur, those involving a temperature change with no change in phase and those at constant temperature with a change in phase.
The heat energy associated with a change in temperature that does not involve a change in phase is given by
A)
Express your answer to three significant figures and include the appropriate units.
B)
Express your answer to three significant figures and include the appropriate units.
gExplanation / Answer
A) Ice (64.0 g, -18C) -------> Water (64.0 g, 25C)
The process consists of three sub-processes:
i) Ice (-18C) -------> Ice (0C) =====> Temperature change with no phase change
q1 = m*s*T = (64.0 g)*(2.09 J/g.C)*[(0C) – (-18C)] = (64.0*2.09*18) J = 2407.68 J = (2407.68 J)*(1 kJ/1000 J) = 2.40768 kJ 2.408 kJ
ii) Ice (0C) ------> Water (0C) ====> Phase change at constant temperature
q2 = m*Hfus = (64.0 g)*(334 J/g) = 21376 J = (21376 J)*(1 kJ/1000 J) = 21.376 kJ
iii) Water (0C) -------> Water (25C) ====> Temperature change with no phase change
q3 = m*s*T = (64.0 g)*(4.18 J/g.C)*(25 – 0)C = (64.0*4.18*25) J = 6688 J = (6688 J)*(1 kJ/1000 J) = 6.688 kJ
The total heat energy required is q = q1 + q2 + q3 = (2.408 + 21.376 + 6.688) kJ = 30.472 kJ (ans)
B) Molar mass of water, H2O = (2*1.008 + 1*15.9994) g mol-1 = 18.0154 g mol-1.
We have 1.50 mole of water; therefore, mass of water present = (1.50 mole)*(18.0154 g mol-1) = 27.0231 g.
The given process is a phase change at constant temperature; the heat energy required is q = m*Hvap = (27.0231 g)*(2250 J/g) = 60801.975 J.
The time required = (60801.975 J)/(24.0 J/s) = 2533.416 s = (2533.416 s)*(1 min/60 s) = 42.2236 min 42.224 min (ans).
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