Asked by Please Help!
If 1.10 g of steam at 100.0 degrees celsius condenses into 38.5 g of water, initially at 27.0 degrees celsius, in an insulated container, what is the final temperature of the entire water sample? Assume no loss of heat into the surroundings
Answers
Answered by
DrBob222
heat lost by 100 C water + heat gained by cold water - heat lost by condensation =0
1.1*4.184*(Tf-100) +[38.5*4.184*(Tf-27)] - 1.1*2257 = 0
Solve for Tf.
You can do it another way if you wish.
First calculate how much the condensing steam will raise the T of cold water.
1.1*2257 = 2482.7 J released.
38.5*4.184*(Tf-27) = 2482.7
Tf = 42.41 THEN
heat lost by 1.1g steam @ 100C + heat gained by 41.41 C water = 0 and recalculate Tf EXCEPT, of course, Tinitial is 42.41 and not 27.0 C. Same answer is obtained.
1.1*4.184*(Tf-100) +[38.5*4.184*(Tf-27)] - 1.1*2257 = 0
Solve for Tf.
You can do it another way if you wish.
First calculate how much the condensing steam will raise the T of cold water.
1.1*2257 = 2482.7 J released.
38.5*4.184*(Tf-27) = 2482.7
Tf = 42.41 THEN
heat lost by 1.1g steam @ 100C + heat gained by 41.41 C water = 0 and recalculate Tf EXCEPT, of course, Tinitial is 42.41 and not 27.0 C. Same answer is obtained.
Answered by
Please Help!
Thank you! what is the 2257 from?
Answered by
Chandler
2257 is the heat of vaporization of water.
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