Final answer:
The final temperature when 32.2 g of water at 14.9 °C mixes with 32.2 grams of water at 46.8 °C is calculated using the principle of heat transfer, and it results in a final temperature of 30.85°C.
Step-by-step explanation:
When two samples of water at different temperatures are mixed, the final temperature of the mixture can be found using the concept of heat transfer. Since water has a specific heat capacity, we can set up the heat gained by the cooler water equal to the heat lost by the warmer water, because no heat is lost to the surroundings. Assuming no heat loss, the equation can be written as:
m1*c*(Tfinal - Tinitial1) = m2*c*(Tfinal - Tinitial2)
Where 'm' represents mass, 'c' is the specific heat capacity of water (4.18 J/g°C), 'Tfinal' is the final temperature we want to find, and 'Tinitial' are the initial temperatures of the two masses of water. Since the masses and specific heat capacities are the same, it simplifies to:
(32.2 g)(4.18 J/g°C)(Tfinal - 14.9°C) = (32.2 g)(4.18 J/g°C)(46.8°C - Tfinal)
Solving for 'Tfinal' gives us a final temperature of 30.85°C.
a