Question

Not sure where to start:
Solve for Tf [1.0][100-Tf][0.50] = [50.0][Tf-20][4.184 ]

Answer 1

Not sure where to start:
\[ Solve for T _{f} [1.0][100-T _{f}][0.50] = [50.0][T _{f}-20][4.184 ]\]

Answer 2

what do you meant by
\[T_f\]
here

Answer 3

Well, I typed it weird. All the info I have is to solve for Tf (with the f being a sub). So the real problem is
\[[1.0][100−Tf][0.50]=[50.0][Tf−20][4.184]\]

Answer 4

what is T and F here then?

Answer 5

\[[1.0][100−T _{f}][0.50]=[50.0][T _{f}−20][4.184]\]

Answer 6

Tf probably stands for Temperature_final

are the square braces really just parentheses?

Answer 7

if we use x for Tf
(100-x) * ½ = 50*4.184 *(x-20)
100-x = 100*4.184(x-20)
100 - x = 481.4x - 20*418.4
add x to both sides
100 = 482.4x - 20*418.4
add 20*418.4 to both sides
100+20*418.4= 482.4 x

divide both sides by 482.4

Answer 8

It doesn't say. Simply says solve for Tf. I think it's just an exercise ... I'm not sure about the square braces. They are there as square braces. Don't know if she means ( ).

Answer 9

are you studying algebra?or something I think It's just asking to solve for Tf

Answer 10

Yes, algebra

Answer 11

DO you expect this ..:
Then here is the answer
\[\frac{ 100 - T_f }{ 2 } = [50T_f - 1000][4.184]\]
\[100-T_f = 2([50T_f -1000] [4.184])\]

carrryibng LHS?RHs working we can work out T_f

Answer 12

Not sure where you got the 1000, Adjax??

Answer 13

The water and the metal will end at the same temperature, Tf (T final)
The heat lost by the metal will be equal to the heat gained by the water.
The heat gained(or lost) = specific heat x mass x temperature change

Heat lost by metal = 0.50 J/gC x 1000 g x (100 - Tf)
(0.50 kJ/kgC = 0.50 J/gC)
Heat gained by water = 4.184 J/gC x 50.0 x (Tf - 20.0)

0.50 x 1000 x (100 - Tf) = 4.184 x 50.0 x (Tf - 20)
50000 - 500Tf = 209.2Tf - 4184
54184 = 709.2 Tf
Tf = 76.4 C

Answer 14

did that help you

Answer 15

Don't see how you got to the 2nd line from the 1st line...

Answer 16

??

Answer 17

A 1.0 kg sample of metal with a specific heat of 0.50 J/g (degrees) C is heated to 100 (degrees) C and then placed in a 50.0 g sample of water at 20.0 (degrees) C. What is the final temperature of the metal and the water?

Answer 18

that shold help you if not chat me

Answer 19

@gra first toy tell us that you are studying algebra/maths ..now here's a solution ABOVE OF PHYSICS WHICH IS ALSO RIght in it's frame of reference..

So for what you are asking solution for?

Answer 20

@cece1902 .. not sure why this is so difficult for me to understand! I think the teacher is simply wanting us to go through the algebraic steps...but I'm having trouble setting it up.

Answer 21

ok just put this.A 1.0 kg sample of metal with a specific heat of 0.50 J/g (degrees) C is heated to 100 (degrees) C and then placed in a 50.0 g sample of water at 20.0 (degrees) C. What is the final temperature of the metal and the water?

Answer 22

did that help you

Answer 23

or not