Question

If you combine 300.0 mL of water at 25.00°C and 110.0 mL of water at 95.00°C, what is the final temperature of the mixture?

Answer 1

I got `44^@C` approx.

Explanation:

We can say that the heat lost by the hot water `Q_1` is equal to the heat `Q_2` gained by the cold where the heat can be written as funcion of mass `m`, specific heat `c` and the variation of temperature between final and initial temperature `DeltaT=(T_f-T_i)`.
We will have:
`Q_1+Q_2=0`
`m_1cancel(c)(T_f-T_1)+m_2cancel(c)(T_f-T_2)=0`

With `1ml=1mg` for mass:

`110(T_f-95)+300(T_f-25)=0`
`110T_f-10,450+300T_f-7,500=0`
`410T_f=17,950`
So:
`T_f=(17,950)/410=43.78~~44^@C`