275g of liquid water at 100'C and 550g water at 30'C are mixed together, what is the final temp

I am so stuck with this question, any help from you chemistry geniuses would be very appreciated!

275g of liquid water at 100'C and 550g water at 30'C are mixed together, what is the final temp?

Heat lost = heat gained.
c = 4.19 J/oC*gram which will cancel out.

275*c*(100 - t) = heat lost. Notice the t is the second term in the temperature.

550*c*(t - 30) = heat gained. Notice the t is the first term in the temperature.

You may wonder why you use 30 and 100 as you do. They are done that way because before they are mixed they have enough energy to be at 30o and 100o The 30 is subtracted from the final temperature because you need to add the heat to go to the final temperature from 30.

For the 100 degrees, you need to lose heat, but only enough to get from 100 to t.

275 * (t - 30) = 550*(100 - t) Remove the brackets and solve for t. It comes out to 76.7 oC

Notice that before I began, I divided by c. It is the same on both sides and so does not effect the result.

This equation was created by Joseph Black
It's used to calculate mixing temperature of two liquid which has different temperature.
(M1 X C1) (T1-Ta) = (M2 X C2) (Ta-T2)

M represent mass
C represent specific heat capacity (depend on temperature)
T represent temperature

Specific heat capacity of water @ 30 C is 4.179 (J/gram.C)
Specific heat capacity of water @ 100 C is 4.219 (J/gram.C)


275 X 4.219 X (100 - Ta) = 550 X 4.179 X (Ta - 30)

And we get Ta (final mixing temperature) is 53.48 C

Formula is
mass x specific heat(tF-ti) = -[mass x specific heat(ti-tf)]