When 131.0 ml of water at 26.0°c is mixed with 81.0 ml of water at 85.0°c, what is the final temperature? (assume that no heat is lost to the surroundings; d of water is 1.00 g/ml.)

63.52°C is the final temperature

Explanation:

1) 131.0 mL of water at 26.0°C

Mass of water = m

Volume of the water =131.0 mL

Density of the water = 1.00 g/mL

m = 131.0 g

Initial temperature of the water = = 26.0°C

Final temperature of the water =

Change in temperature ,

Heat absorbed 131.0 g of water = Q

2) 81.0 mL of water at 85.0°C

Mass of water = m'

Volume of the water =81.0 mL

Density of the water = 1.00 g/mL

m' = 81.0 g

Initial temperature of the water = = 85.0°C

Final temperature of the water =

Change in temperature ,

Heat lost by 81.0 g of water = Q'

After mixing both liquids the final temperature will become equal fro both liquids.

Since, heat lost by the water at higher temperature will be equal to heat absorbed by the water at lower temperature.

Q=-Q' (Law of conservation of energy.)

Let the specific heat of water be c

63.52°C is the final temperature

explanation: