need to know how to get the resulting temperature on this problem?

Question

you dont have to give me the answer. just the formula on how to do this would be enough. im really lost. any help would be appriciated. so here it is.

what is the resulting temperature if 6000 grams of water at 60'C is mixed with 400 grams water at 100'C in a container of neglible thermal capacity?

thanks....

Answer 1

Assume that the heat capacity of water is x (because it doesn't matter what the real heat capacity is) then you have

6000*60*x + 400*100*x = 6400*T*x

Drop the x, then solve for T

Answer 2

This problem is relatively simple sine we are dealing with the same substance thus eliminating one variable, i.e. specific heat.

m.s.t = Q1 (Heat capacity) where m is the mass, s the specific heat and t the temperature of an object.

Here we have 6000.1.60 for the first mass and 400.1.100 for the second mass. Specific heat of water is 1.

Adding them together we get 360000 + 40000 = 400000 calories

Temperature of the mixture will be 400000/6400 = 62.5 C

If the container's thermal capacity is not to be neglected, the temperature will be lower than this value.

Answer 3

Here we have 3 mass figures:
M1 = 6.0kg: M2 = 0.400kg: and M3 = 6.400kg
We also have 3 temperature figures
T1 = 60°C: T2 = 100°C: T3 = ?
T3 = (M1 x T1) + (M2 x T2) ÷ M3
T3 = (6.0 x 60) + (0.40 x 100) ÷ 6.400