Question

If you combine 400.0 mL of water at 25.00 �C and 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.

Answer 1

Let the final temperature of the mixture be t^oC   ( Obviously, 95.00 ^oC > t ^oC > 25.00 ^oC )

Mass of the sample of cold water = ( Volume x Density ) = ( 400 mL x 1.00 g/mL ) = 400 g

Specific heat capacity of water = 4.184 J/g. ^oC

Heat gained by cold water = Mass x Specific heat capacity of water x Change in temperature

                                          = (400g ) x ( 4.184 J/g. ^oC ) x ( t - 25) ^oC

Mass of the sample of hot water = ( 140 mL x 1.00 g/mL) = 140 g

Heat lost by hot water = ( 140 g) x ( 4.184 J/g. ^oC) x ( 95 - t) ^oC

According to the principle of calorimetry,

Heat gained = Heat lost

(400 g) x ( 4.184 J/g. ^oC) x ( t - 25) ^oC = (140 g) x ( 4.184 J/g. ^oC) x ( 95 - t ) ^oC

400 ( t - 25 ) = 140 ( 95 - t)

400 t - 10000 = 13300 - 140t

540t = 23300

t = 43.15

Therefore, the final temperature of the mixture is 43.15 ^oC.