A 350g glass beaker contains 500g of water at a temperature of 10.0°C. If 400g of ethyl alcohol at 35°C is subsequently poured into
The beaker and the water and alcohol mixture is thoroughly stirred, what is its equilibrium temperature?

The heat absorbed by the glass and water must equal the heat given up by the alcohol as it all comes to equilibrium temperature. If the final temperature rise of the water and glass is \Delta T, then the temperature drop of the alcohol is 25 - \Delta T. So you have:


(350 g \times C_{glass} \ + \ 500g \times C_{water}) \Delta T = 400g \times C_{alcohol} \times (25 - \Delta T )


Presumably you have values for the various specific heats of each of these substances - solve for \Delta T. Then the final temp is 10 + \Delta T.