In: Chemistry
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.
Let the final temperature of the mixture be toC ( 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.