If you combine 320.0 ml of water at 25.00 °c and 120.0 ml of water at 95.00 °c, what is the final temperature of the mixture? use 1.00 g/ml as the density of water.

The heat from the hotter water will go into the colder water untl equilibrium is reached. Equilibrium is same temperature!

Now, the heat is proportional to the mass, the specific heat and the temperature difference. The specific heat does not matter since all is water, it will cancel out:

m_1 * c_H20 * ( T_final - T_1 ) = -m_2 * c_H20 * ( T_final - T_2)

Notice the minus, because one wins the heat of the one who loses it. In this way both sides have the same sign:

m_1*(T_final - T_1)=-m_2*(T_final-T_2), or after some simple algebra:

T_final = (m_1 * T_1 + m_2 * T_2 )/(m_1+m_2),

which looks like an arithmetic mean, and one could have gone for this, but the above shows all the work. Notice that if T_1=T_2, T_final=T_1 always, which makes sense.

Now you can convert volume to mass with the density, but since mass = density*volume and it is all water, the density will cancel out and you can work with volumes. If you prefer just say: 120 ml->120 g , etc ...

T_final = (120*95+320*25)/(320+120)=44.0909 degrees Celsius, or ~ 44.09 degrees with two decimal precision as your statement (beware of precision always!).

Sorry i dont understand the question can you rephase it?

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