Question

Confusion on transfer of heat formula..

"Calculate the equilibrium temperature when a 5 gram sample of lead at 20 degrees celsius is placed in contact with a 10 gram block of silver at 40 degrees celsius.

Answer 1

@Abhisar

Answer 2

\[q=c_p \times m \times \left| \Delta T_C \right|_{Lost}=C_p \times m \times \left| \Delta T_c \right|_{gained}\]
This is the formula i am using

Answer 3

So I gather all the specific heats, mass, etc and I get

(0.235j/g)(10g)(40-tfinal)=(0.140j/g)(5g)(tfinal-20)

then im not sure what to do..

Answer 4

@Jhannybean

Answer 5

@butterflydreamer

Answer 6

@alias

Answer 7

equate the two heat equations of the two metals together

Answer 8

thats what i did lol

Answer 9

(0.235j/g)(10g)(40-tfinal)=(0.140j/g)(5g)(tfinal-20)

Answer 10

oh lol sorry didn't see. Well don't you just simplify and solve for t-final?

Answer 11

how should i simplify it? multiply out the specific heats and grams?

Answer 12

yeah, the 0.235 and the 0.14 is the specific heat right? In that case you've made a mistake in the unit, it should be joules per gram PER CENTIGRADE, you're missing the Celsius.

But yea, multiply out everything, it's just simple algebra at this point

Answer 13

for some reason, i don't get it...

0.235 times 10=2.35

do i distribute that with tfinal?

like 2.35(40-tfinal)

94-2.35 final..?

Answer 14

@abb0t

Answer 15

yea you do, then you collect the t-final like terms, isolate it on one side, and divide to get t-final = some number

Answer 16

multiply and distribute EVERYTHING, then isolate t-final

It might be easier to see if you replace t-final with x, so that it looks like math I guess...

Answer 17

@Jhannybean

Answer 18

lol yeah. I just replaced tfinal with x and now i understand it..

basic algebra huh

tfinal=36

Answer 19

looks really easy now. thank you dude.

well, it's 3:17 am. off to watch a movie or something.

thanks again