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?

Answer 1

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.

Answer 2

150.0g x 4.184x(85.5)= 53659.80
+
1000ml x 4.184x 5 = 20920

74579.80 / 4.184=

I DON'T GET IT!

Answer 3

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.

Answer 4

BTW, thanks for showing your work. That makes it easy to spot what's wrong.

Answer 5

You should get an answer that's about 79 C for Tfinal.

Answer 6

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.

Answer 7

Verve the vrbt