Question

if you combine 420 mL of water at 25 degrees C and 130 mL of water at 95 degrees C what is the final temperature of the mixture

Answer 1

There are several important information's already given in the question. Based on those information's the answer can be easily deduced.
We already know that
Heat gained + Heat lost = 0
Also
Q = mass x specific heat x delta T
So
420 mL x 1 g/mL = 420 g
130 mL x 1 g/mL = 130 g
Now
420( T - 25 ) + 130 ( T - 95) = 0
420T - 10500 + 130T - 12350 = 0
550T - 22850 = 0
550T = 22850
T = 41.5 degree centigrade
I hope that the procedure is clear enough for you to understand and the answer has come to your great help.