Question

A 48 g sample of water at 98.°C is poured into a 55 g sample of water at 15°C. What will be the final temperature of the mixture?

Answer 1

recall that the amount of heat absorbed or released is given by:
Q = mc(T2-T1)
where
m = mass (in g)
c = specific heat capacity (in J/g-K)
T = temperature (in C or K)
*note: Q = (+) when heat is absorbed and (-) when heat is released

to get the final temp, we note that Q,released = Q,absorbed. therefore:
Q,released = Q,absorbed
-m1*c*(T2-T,a1) = m2*c*(T2-T,r1)
since both substances involved are the same (which is water), c is cancelled:
-m1*(T2-T,a1) = m2*(T2-T,r1)
-48*(T2 - 98) = 55*(T2 - 15)
solving for T2,
T2 = 53.68 C

hope this helps~ :)