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Chemistry phase change question - ice added to steam?
75.0 g of ice at -10.0C is added to 10.0 g of steam at 100C. What is the final temperature of the 85.0 g of water? (specific heat of ice water = 2.03 J/g-C; gas = 2.00 J/g-C; heat of fusion = 6.02 kJ/mol; heat of vaporization = 43.9 kJ/mol)
I'm not really sure how to start this problem because I don't know when to use mass = 75 g and when to use 85 g. I know the ice has to go from -10C to 0C and then change to liquid, but not sure how to incorporate the steam. The final answer is supposed to be 5.5C. Any help would be much appreciated!
Thank you so much! That makes sense. My only question is in the steam's 2nd step why, for delta T, you wrote (100-Tf). The math works but why isn't it (Tf - 100)??
2 Answers
- HPVLv 79 years agoFavorite Answer
Treat the ice and steam as if they are two different substances. The amount of heat gained by the ice is the same as the amount of heat lost by the steam.
Note: heat of fusion of ice = 6.02 kJ/mole x (1 mole H2O / 18.0 g H2O) = 0.334 kJ/g = 334 J/g
Note: heat of vaporization of steam = 43.9 kJ/mole x (1 mole H2O / 18.0 g) = 2.44 kJ/g = 2440 J/g
Heat gained by the ice: the ice will heat up in 3 steps: (1) goes from H2O(s) at -10 C to H2O(s) at 0 C, (2) melts at 0 C, and (3) again heats up as H2O(l)
Step 1: Heat gained = (mass ice)(sp. ht. ice)(Tf - Ti) = (75.0 g)(2.03 J/g C)(0 C - (-10 C) = 1523 J
Step 2: Heat gained = (mass ice)(heat of fusion of ice) = (75.0 g)(334 J/g) = 25,050 J
Step 3: Heat gained = (mass H2O(l))(sp. ht. H2O(l))(Tf - Ti) = (75.0 g)(4.18 J/g C)(Tf - 0 C) = 314Tf
Total heat gained = 1523 J + 25,050 J + 314 Tf = 26,570 + 314Tf
Heat lost by steam: will occur in 3 steps: (1) the steam condense to H2O(l) at 100 C), and (2) the H2O(l) will then cool down to Tf.
Step 1: Heat lost = (mass steam)(heat of condensation of steam) = (10.0 g)(2440 J/g) = 24,400 J
Step 2: Heat lost = (mass H2O(l))(sp. ht. H2O(l))(Ti - Tf) = (10.0 g)(4.18 J/g C)(100 C - Tf) = 4180 - 41.8 Tf
Total heat lost = 24,400 J + (4180 - 41.8Tf) = 28,600 - 41.8Tf
So heat gained by ice = heat lost by steam
26570 + 314Tf = 28600 - 41.8Tf
356Tf = 2030
Tf = 5.7 C . . .close to 5.5 C
check my math
- Anonymous4 years ago
a million) H2O(s) -18.0 C --> H2O(s) at 0.0 C --> H2O(l) 0.0 C --> H2O(l) 25.0 C q1 = m.s.Delta T = (75.0g)(2.09J/g.C)(18.0 C) = --------- J q2 = g.Delta Hfus= (75.0g)(334 J/g)= --------- J q3 = m.s.Delta T = (75.0g)(4.18 J/g/C)(25.0C) = ________ J upload q1 + q2 + q3 to get the reply. 2) q = g.Delta Hvap = (a million.50 mol)(18.0g/1mole)(2250J/g) = ________ J __________ J (1s/15.0J)(1min/60s) = _________ min because that i do no longer have the calculator with me, I have purely shown you the setups. Do the mathematics yorself.
