Question

if 100.0 g of water at 90 deg. C is added to 50.0 g of water at 10 deg. C, estimate the final temperature of the water. I think this should be 50 deg. C, but when I tried to see the change in both, the cold one is 40, the hot one is 50. Both of the changes should be the same.

Answer 1

\[q=c_s*m*\Delta T\] I'm sure you are going to need that one. I'm having trouble figuring this out though. C=4.184

Answer 2

If the amounts of water added were the same, we could expect the same temperature change. Since you're adding a little bit of cold water to a lot of hot water, the change won't be split down the middle. The final temp will be the same, so this makes the equation slightly more complicated:
\[Q{_h}{_o}{_t} = m{_h}{_o}{_t}*C*(T{_f}{_i}{_n} - T{_h}{_o}{_t})\] and
\[Q{_c}{_o}{_l}{_d} = m{_c}{_o}{_l}{_d}*C*(T{_f}{_i}{_n} - T{_c}{_o}{_l}{_d})\]
the laws of thermo say that whatever heat is lost by one portion of a system must be gained by the other, so we can set those 2 equations equal to each other!\[-m{_h}*C*(T{_f} - T{_h}) = m{_c}*C*(T{_f} - T{_c})\]we have to add a negative to one side, because the change in temp from of the hot water will be negative, so we have to cancel it. Since both liquids are water, we don't even ned the value of C, because it cancels also. The only piece you don't know is Tf, so rearrange and solve

Answer 3

oh, thank you. Btw what is

Answer 4

TF is the final temperature of the mixture, and the thing you're trying to solve for.

Answer 5

is m the mass?

Answer 6

yes the m is mass

Answer 7

i got like -189 sth, it's not right answer, can s1 help me?