Asked by RoseBud

what will be the final temperature of a system in which 150.0g of water at 5.0 degrees C are added to 1.00L of water at 90.5 degrees C?
Answered by DrBob222
heat absorbed by cool water + heat lost by warm water = 0

[mass cool water x specific heat water x (Tfinal-Tinitial)] + [mass warm water x specific heat water x (Tfinal-Tinitial)] = 0
Solve for Tfinal.
Answered by RoseBud
150.0g x 4.184x(85.5)= 53659.80
+
1000ml x 4.184x 5 = 20920

74579.80 / 4.184=


I DON'T GET IT!
Answered by DrBob222
You didn't substitute correctly.
It should be
[150.0 x 4.183 x (Tf-5)] + [1000 x 4.184 x (Tf-90.5)] = 0 and you solve for Tf.
Answered by DrBob222
BTW, thanks for showing your work. That makes it easy to spot what's wrong.
Answered by DrBob222
You should get an answer that's about 79 C for Tfinal.
Answered by RoseBud
This makes sense. Thank you so much. This a great website. I don't just want someone to tell me the answer....I want to know how to figure it out. Thanks.
Answered by Anonymous
Verve the vrbt
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