If you combine 270.0 mL of water at 25.00 °C and 100.0 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.

Answer 1

The general principle is conservation of energy. Accordingly, the heat LOST by the hot water must equal the heat GAINED by the cooler water.

Heat lost by hot water = q = mC∆T

q = heat = ?

m = mass = 100 ml x 1 g / ml = 100 g

C = specific heat of water = 4.184 J / gº

∆T = change in temperature of the hot water = (95.00º - Tf) where Tf is the final temperature

q = (100 g)(4.184 J/gº)(95 - Tf)

q = 418.4 (95 - Tf)

q = 39748 - 418.4Tf = heat lost by hot water

Heat gained by cooler water = q = mC∆T

q = heat

m = mass = 270 ml x 1 g / ml = 270 g

C = specific heat = 4.184 J/gº

∆T = change in temperature of the cooler water = Tf - 25º

q = (270 g)(4.184 J/gº)(Tf - 25)

q = 1129.7Tf - 28242 = heat gained by cooler water

Setting heat lost = heat gained, we have ....

39748 - 418.4Tf = 1129.7Tf - 28242

1548.1Tf = 67990

Tf = 43.92º