Final Temperature Calculation for Water Mixing in Insulated Container

[Rhine720]

Homework Statement

if 100.0g of water at 100.0 C is mixed with 200g of water at 20.0 in an insulated container, what will the final temperature be?

Homework Equations

Q=MCdeltaT

The Attempt at a Solution

Since the energy change would be equal but opposite I though oh well I can leave the Q out and have (100.0)4.18(Tf-100.0)=(200)4.18(Tf-20.0). So I sort of did worked through this but started getting crazy stupid answer that I threw away. Pretty much simplified,distributed,addition and subtraction property of equality an then division property of equality

 

[Delphi51]

(100.0)4.18(Tf-100.0)=(200)4.18(Tf-20.0)

is correct except for trouble with signs: the left side is negative and the right side is positive. Replace the Tf - 100 with 100 - Tf.

 

[Rhine720]

is correct except for trouble with signs: the left side is negative and the right side is positive. Replace the Tf - 100 with 100 - Tf.

Thanks.. and then

418x(100.0-Tf)=836(Tf-20.0)
41800-418Tf=836Tf-16700(in sig figs)
58500=1254Tf
46.7C(sig fig)

?

 

[Delphi51]

Looks good!