If you have 320.0 mL of water at 25.00 °C and add 140.0 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 sum of the heats gained is zero (one looses heat, so is a negative heat gained).
Heatgaindcool+heatgainedhot=0
320c(Tf-25)+140c(tf-95)=0
do the algebra (notice c divides out), solve for Tf
how do i solve for Tf?
To find the final temperature of the mixture, you can use the principle of conservation of energy. The total energy of the system remains constant before and after mixing.
To solve this problem, you can use the formula:
(m1 × c1 × ΔT1) + (m2 × c2 × ΔT2) = 0
Where:
m1 and m2 are the masses of the two samples of water
c1 and c2 are the specific heat capacities of water (approximately 4.18 J/g·°C)
ΔT1 and ΔT2 are the changes in temperature for the two samples of water
First, let's calculate the masses of the two samples of water. We can use the density of water, which is given as 1.00 g/mL:
mass1 = volume1 × density = 320.0 mL × 1.00 g/mL = 320.0 g
mass2 = volume2 × density = 140.0 mL × 1.00 g/mL = 140.0 g
Now we can plug in the values into the formula to find the final temperature:
(320.0 g × 4.18 J/g·°C × (Tf - 25.00 °C)) + (140.0 g × 4.18 J/g·°C × (Tf - 95.00 °C)) = 0
Using this equation, we can solve for Tf, the final temperature of the mixture.