Using the concept of the equilibrium temperature, calculate what your resultant temperature.

i had two cups of water - one 16oC and one 84oC. they both were 250mL when i mixed them together the temperature was 45oC, I am being asked to Use the concept of the equilibrium temperature, calculate what your resultant temperature should be for your mixed water. Assume insulation from the surroundings for this calculation. Then i need to compare the result to my finding of 45oC.The formular i have been given is this - 0 = 饾憵1饾憪1螖饾憞1 + 饾憵2饾憪2螖饾憞2
I am so lost and confused. any help would be appreciated.

Thanks

I can't really see your formula but I will assume it is something like:
m1 C1 ( Tf - T1 ) + m2 C2 ( Tf - T2 ) = 0

This is a heat balance equation where,
m1 = mass of water #1 (density=mass/volume, mass = 999 kg/m3 x 2.50 * 10^-4 m3 = 0.250kg)
m2 = mass of water #2 (mass = 972 kg/m3 x 2.50 * 10^-4 m3 = 0.243kg)
C1 = Specific heat capacity of m1 (in this case, water = 4.18 J/(g.K))
C2 = Specific heat capacity of m2 (also 4.18 J/(g.K))
T1 = Initial temperature of m1 (16 degrees Celsius = 289.15K )
T2 = Initial temperature of m2 (84 degrees = 357.15K)
Tf = Final temperature of mixture

So plugging all these into the equation and solving for Tf,
250g x 4.18 J/(g.K) (Tf - 289.15)K + 243g x 4.18 J/(g.K) (Tf - 357.15)K = 0
1045 Tf - 302161.75 + 1015.74 Tf - 362771.54 = 0
Tf = 322.67 Kelvins = 49.67 degrees Celsius

This result is 4.67 degrees higher than the resultant temperature because heat is lost to the surroundings. The calculations do not take this into account.

A little tedious but you really just have to watch out for the units in these types of questions.

Hope it's right.