Question

a 100g sample of wate at 90oC is added to a 500g sample of water at 10oC. what is the final temperature of the water

Answer 1

the heat released or absorbed is given by the formula,
Q = mc(T2-T1)
where
m = mass of substance (units in g)
c = specific heat capacity (units in J/g-K)
T2 = final temperature
T1 = initial temperature
**note: if Q is (-), heat is released and if (+), heat is absorbed

therefore, we can write,
Q,absorbed = Q,released
Q,a = -Q,b
m1*c1*(T2 - T1,a) = -m2*c2*(T2 - T1,b)
since both substances are water, c1 = c2, and we can cancel them out. substituting the values,
m1*(T2 - T1,a) = -m2*(T2 - T1,b)
100*(T2 - 90) = -500*(T2 - 10)
100*T2 - 9000 = -500*T2 + 5000
600*T2 = 14000
T2 = 23.33 degree Celsius

note: final temperature, T2, is always the same for both (since it is the equilibrium temperature). also, T2 must be always between the initial temperatures of the substances - may be close but never equal.

hope this helps~ :)