Question

how do you do a problem like...
"If a 10.0g ball of iron at 160.0c is dropped into 50.0g of water at 20.0c in an insulated container, what will be the final temp of the water? the specific hear of iron is .444j/g and that of water is 4.18j/g."?

Answer 1

M1.Ciron.T1\[M iron \times C (calor specific of iron) \times T1 ^{2} = M water \times C ( calor specific of water) \times T 2^{2}\]

Answer 2

black rules CMIIW

Answer 3

M= mass (kg)

Answer 4

what are the variables in that tough?

Answer 5

black rules please googling

Answer 6

because as i see that i would do
10*.444*160^2 = 50*20*4.18

Answer 7

oooh sorrryyyy i made mistakes

Answer 8

Q=mc(T2-T1)

Answer 9

would i have to use that twice ? or is there a way i can set it up to find it in one step?

Answer 10

sorry the message board is laggy, since the conatiner is insulated solve Q= 0 = Qwater + Qiron in which Qwater=mwater x cwater (T2-Twater) and Qiron=miron*ciron(T2-Tiron) the T2 is the same variable, basically plug numbers in and solve Qiron=-Qwater isolate T2 and your finished

Answer 11

i'm off for lunch, good luck!

Answer 12

kk ty