# What is the final temperature when 100 g of water at 75"^oC is mixed with 75 g of water at 80"^oC?

Dec 30, 2017

These two systems will equilibrate. The system at a higher initial temperature will be exothermic, transferring heat to the contextually endothermic system.

Hence,

-75g * (4.18J)/(g*°C) * (80°C - T_f) = 100g * (4.18J)/(g*°C) * (75°C - T_f)

therefore T_f approx 77°C

To be sure, this is but a difficult representation of the familiar scenario,

$q = m {C}_{s} \Delta T$,

where the heat supplied is merely coming from the exothermic system.